The pressure of water on ground is 50000pa and at first floor it is 20,000pa .find the height of the first floor??​

Answers

Answer 1

The height of the first floor is 7.5 meters if the water pressure on the ground is 50000 pa and 20,000 pa at the first floor.

How is the height of the first floor determined?

Using the hydrostatic pressure equation, we can get the reference level as the water pressure at the ground floor:

P = ρgh

P is equal to 50000 Pa on the ground floor and 20000 Pa on the first. Water's constant density allows us to write:

P1/P2 = h1/h2

where P1 and h1 represent the ground floor pressure and height and P2 and h2 represent the first floor pressure and height.

Inputting the values provided yields:

50000/20000 = h1/h2

As a result, the first level is 2.5 times as tall as the bottom floor. The height of the first floor would be as follows if we used a typical height of 3 meters per storey:

2.5 × 3 = 7.5 meters for h2.

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Related Questions

A 0. 068-kg ball strikes a wall with a velocity of 22. 1 m/s. The wall stops the ball in 0. 63 s. What is the magnitude of the
force applied by the wall on the ball?

a. 5. 3n
b. 4. 2n
c. 12n
d. 2. 4n

Answers

The correct answer is (d) 2.4 N. We can use the impulse-momentum theorem to solve this problem. The impulse of the force is equal to the change in momentum of the ball.

The initial momentum of the ball is: p1 = mv = (0.068 kg)(22.1 m/s) = 1.5038 kg*m/s

Since the wall stops the ball, the final momentum of the ball is zero: p2 = 0 kg*m/s

The change in momentum is: Δp = p2 - p1 = -1.5038 kg*m/s

The time interval for the force to act is 0.63 s.

So, the magnitude of the force applied by the wall on the ball is: F = Δp / Δt = (-1.5038 kg*m/s) / (0.63 s) ≈ 2.4 N

Therefore, the correct answer is (d) 2.4 N.

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A hoop (i=mr^2) of radius 0.50 m and a mass of 0.20 kg is released from rest and allowed to o go roll down an inclined plane. how fast is it moving after dropping a vertical distance of 3.0 m?
a. 7.7 m/s
c. 5.4 m/s
b. 6.2 m/s
d 3.8 m/s

Answers

The movement of a hoop has converted potential energy to kinetic energy. The hoop dropped vertically for a distance of 3.0 m and is now moving at a velocity of 7.7 m/s. Therefore, the correct answer is option A.

To determine the velocity of a hoop of mass 0.20 kg and radius 0.50 m after it has fallen a vertical distance of 3.0 m, we can use the principle of conservation of energy.

At the top of the incline, the hoop has potential energy given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height of the incline.

At the bottom of the incline, all of the potential energy has been converted to kinetic energy given by [tex]1/2mv^2[/tex], where v is the velocity of the hoop.

Using conservation of energy, we can set the initial potential energy equal to the final kinetic energy and solve for v. The potential energy at the top of the incline is mgh = [tex](0.20 \;kg)(9.81 \;m/s^2)(3.0 \;m)[/tex] = 5.89 J.

The kinetic energy at the bottom of the incline is [tex]1/2\;mv^2[/tex], so [tex]1/2(0.20 \;kg)v^2 = 5.89 J[/tex]. Solving for v, we get v = 7.7 m/s.

Therefore, the hoop is moving at a velocity of 7.7 m/s after dropping a vertical distance of 3.0 m. This demonstrates the conversion of potential energy to kinetic energy and the use of conservation of energy in solving physics problems. Therefore, the correct answer is option A.

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A beam of light travels from air into a transparent material. the angle of incidence is 25 and the angle of refraction is 17. what is the index of refraction of the material?

Answers

The index of refraction of the transparent material is approximately 1.46.

The index of refraction (n) of a transparent material is defined as the ratio of the speed of light in vacuum to the speed of light in the material. The relationship between the angles of incidence (θ₁) and refraction (θ₂) and the indices of refraction of the two media can be described by Snell's law, which states that:

n₁ sin(θ₁) = n₂ sin(θ₂)

where n₁ is the index of refraction of the first medium (in this case, air), and n₂ is the index of refraction of the second medium (the transparent material).

Given that the angle of incidence is 25 degrees and the angle of refraction is 17 degrees, we can use Snell's law to solve for n₂:

n₁ sin(θ₁) = n₂ sin(θ₂)

(1.000 sin 25°) = n₂ sin 17°

Solving for n₂, we get:

n₂ = (1.000 sin 25°) / sin 17°

n₂ ≈ 1.46

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An electromagnetic wave traveling through a vacuum
has a wavelength of 1.5 × 10–1

meter. What is the

period of this electromagnetic wave?

Answers

The period of the electromagnetic wave is 5×10⁻¹⁰ seconds

What is period?

Period is the time taken for a wave to complete one rotation.

To calculate the period of the wave, we use the formula below.

Formula:

T = λ/v...........................Equation 1

Where:

T = Period of the electromagnetic  wavev = Speed of the electromagnetic waveλ = Wavelength of the electromagnetic  wave

From the question,

Given:

λ = 1.5×10⁻¹ mv = 3×10⁸ m/s

substitute these values equation 1

T = 1.5×10⁻¹ /3×10⁸T = 5×10⁻¹⁰ seconds

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PROBLEM SOLVING


1. An electron is traveling to the north with a speed of 3. 5 x 106 m/s when a magnetic field is turned on. The strength of the magnetic field is 0. 030 T, and it is directed to the left. What will be the direction and magnitude of the magnetic force?



2. The Earth's magnetic field is approximately 5. 9 × 10-5 T. If an electron is travelling perpendicular to the field at 2. 0 × 105 m/s, what is the magnetic force on the electron?



3. A charged particle of q=4μC moves through a uniform magnetic field of B=100 F with velocity 2 x 103 m/s. The angle between 30o. Find the magnitude of the force acting on the charge.



4. A circular loop of area 5 x 10-2m2 rotates in a uniform magnetic field of 0. 2 T. If the loop rotates about its diameter which is perpendicular to the magnetic field, what will be the magnetic flux?

Answers

The magnitude of the magnetic force on the electron is 1.47 x 10⁻¹⁴ N, directed toward the west.

The magnitude of the magnetic force on the electron is 1.88 x 10⁻¹⁴ N.

1. The direction of the magnetic force on the electron can be found using the right-hand rule. If the electron is moving north and the magnetic field is directed to the left, then the force will be directed toward the west. The magnitude of the magnetic force can be calculated using the formula F = qvBsinθ, where q is the charge of the electron, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the angle is 90 degrees (since the velocity and magnetic field are perpendicular), so sinθ = 1. Plugging in the values, we get:

F = (1.6 x 10⁻¹⁹ C)(3.5 x 10⁶ m/s)(0.030 T)(1)

  = 1.47 x 10⁻¹⁴ N

As a result, the magnetic field on the electron is 1.47 x 10⁻¹⁴ N and is directed westward.

2. The magnetic force on the electron can be calculated using the same formula as above, F = qvBsinθ. In this case, the velocity of the electron is perpendicular to the magnetic field, so θ = 90 degrees and sinθ = 1. Plugging in the values, we get:

F = (1.6 x 10⁻¹⁹ C)(2.0 x 10⁵ m/s)(5.9 x 10⁻⁵ T)(1)

  = 1.88 x 10⁻¹⁴ N

As a result, the magnetic force on the electron is 1.88 x 10⁻¹⁴ N.

3. The magnitude of the force on the charged particle can be calculated using the formula F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the angle is 30 degrees, so sinθ = 0.5. Plugging in the values, we get:

F = (4 x 10⁻⁶ C)(2 x 10³ m/s)(100 T)(0.5)

   = 4 x 10⁻¹ N

Therefore, the magnitude of the force on the charged particle is 0.4 N.

4. The magnetic flux through the loop can be calculated using the formula Φ = BAcosθ, where B is the strength of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.

In this case, the magnetic field is perpendicular to the plane of the loop, so θ = 90 degrees and cosθ = 0. Plugging in the values, we get:

Φ = (0.2 T)(5 x 10⁻² m²)(0)

    = 0

Therefore, the magnetic flux through the loop is zero.

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The magnitude of the magnetic force on the electron is 1.47 x 10⁻¹⁴ N, directed toward the west.

The magnitude of the magnetic force on the electron is 1.88 x 10⁻¹⁴ N.

What is  Magnetic field?

A magnetic field is a force field that surrounds a magnet or a current-carrying conductor. It is a field of force that affects the behavior of charged particles, such as electrons and protons, and other magnetic materials in the vicinity of the magnet or conductor.

1. The direction of the magnetic force on the electron can be found using the right-hand rule. If the electron is moving north and the magnetic field is directed to the left, then the force will be directed toward the west. The magnitude of the magnetic force can be calculated using the formula F = qvBsinθ, where q is the charge of the electron, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the angle is 90 degrees (since the velocity and magnetic field are perpendicular), so sinθ = 1. Plugging in the values, we get:

F = (1.6 x 10⁻¹⁹ C)(3.5 x 10⁶ m/s)(0.030 T)(1)

 = 1.47 x 10⁻¹⁴ N

As a result, the magnetic field on the electron is 1.47 x 10⁻¹⁴ N and is directed westward.

2. The magnetic force on the electron can be calculated using the same formula as above, F = qvBsinθ. In this case, the velocity of the electron is perpendicular to the magnetic field, so θ = 90 degrees and sinθ = 1. Plugging in the values, we get:

F = (1.6 x 10⁻¹⁹ C)(2.0 x 10⁵ m/s)(5.9 x 10⁻⁵ T)(1)

 = 1.88 x 10⁻¹⁴ N

As a result, the magnetic force on the electron is 1.88 x 10⁻¹⁴ N.

3. The magnitude of the force on the charged particle can be calculated using the formula F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the angle is 30 degrees, so sinθ = 0.5. Plugging in the values, we get:

F = (4 x 10⁻⁶ C)(2 x 10³ m/s)(100 T)(0.5)

  = 4 x 10⁻¹ N

Therefore, the magnitude of the force on the charged particle is 0.4 N.

4. The magnetic flux through the loop can be calculated using the formula Φ = BAcosθ, where B is the strength of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.

In this case, the magnetic field is perpendicular to the plane of the loop, so θ = 90 degrees and cosθ = 0. Plugging in the values, we get:

Φ = (0.2 T)(5 x 10⁻² m²)(0)

   = 0

Therefore, the magnetic flux through the loop is zero.

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Water has a specific heat capacity of 4. 184 J/g℃. The experiment heated 200g of water from 30℃ to 100℃. How much energy is absorbed by the water?

1) 25,104J
2) 83,680J
3) 14,000J
4) 58,576J

Answers

The amount of energy absorbed by the water is 58,576J. The answer is 4) 58,576J.

The formula to calculate the amount of energy absorbed by the water is:

Q = m x c x ΔT

Where Q is the amount of energy absorbed (in Joules), m is the mass of water (in grams), c is the specific heat capacity of water (in J/g℃), and ΔT is the change in temperature (in ℃).

Substituting the given values, we get:

Q = 200g x 4.184 J/g℃ x (100℃ - 30℃)
Q = 200g x 4.184 J/g℃ x 70℃
Q = 58,576J

Therefore, the amount of energy absorbed by the water is 58,576J. The answer is 4) 58,576J.

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What was King Louis XVI's goal for Jacques-Louis David's Oath of the Horatil, 1784

1) to send a moral message

2) to educate the public about antiquity

3) to discourage a revolution

4) to decorate his palace


Number 3 is wrong

Answers

The correct answer is option 1: "to send a moral message." King Louis XVI's goal for Jacques-Louis David's Oath of the Horatii was to promote patriotic values and discourage individualism, and the painting was intended to send a moral message about the importance of loyalty to the state and self-sacrifice.

a parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. the capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) a voltmeter reads 51.0 v when placed across the capacitor. when a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 12.1 v. what is the dielectric constant of the material?

Answers

The dielectric constant of the material is 3.38.

The capacitance of a parallel-plate capacitor with air between its plates is given by:

C = ε0 A / d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

When a dielectric is inserted between the plates, the capacitance increases according to:

C' = k ε0 A / d, where k is the dielectric constant of the material.

From the given information, we can use the equation:

C' = V / Q, where V is the potential difference across the plates and Q is the charge on the plates. Initially, when there is air between the plates, the potential difference is 51.0 V. When the dielectric is inserted, the potential difference drops to 12.1 V, but the charge on the plates remains the same.

Therefore, we can write:

C' = V / Q = 12.1 V / Q = k (51.0 V / Q) = 51.0 k / C,

where C is the initial capacitance (with air between the plates).

Solving for k, we get:

k = C' / C = (12.1 V / Q) / (51.0 V / Q) = 0.2373.

Using the equation for the capacitance with a dielectric, we can also write:

C' = k ε0 A / d,

which gives us:

k = C' d / (ε0 A) = 3.38.

As a result, the material's dielectric constant is 3.38.

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Galileo is often credited with the early discovery of four of Jupiter's many moons. The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the sun. One of the moons is called Io - its distance from Jupiter's center is 4. 2 gigameters and it orbits Jupiter in 1. 8 Earth-days. Another moon is called Ganymede; it is 10. 7 gigameters from Jupiter's center. What is Ganymede's period in Earth days?

Answers

Ganymede's period in Earth days is approximately 7.16 days.

The period of Ganymede in Earth days can be calculated using Kepler's Third Law of Planetary Motion, which states that the square of a planet's period (in Earth days) is proportional to the cube of its average distance from the center of its orbit. Mathematically, this can be represented as:

(T1^2/T2^2) = (R1^3/R2^3)

Where T1 and T2 are the periods of Io and Ganymede respectively, and R1 and R2 are their distances from Jupiter's center. Substituting the given values for Io and Ganymede, we get:

(1.8²/T2²) = (4.2³/10.7³)

Solving for T2, we get:

T2 = 7.16 Earth-days

As a result, Ganymede's period on Earth is around 7.16 days.

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The total mass of the cart is 1. 00 kg, and the mass that is hanging is 0. 200 kg. Calculate the net force on the system, then the acceleration of the system

Answers

The total mass of the cart is 1. 00 kg, and the mass that is hanging is 0. 200 kg.

1. To calculate the net force on the system, we need to consider the forces acting on both masses. The mass hanging from the pulley experiences a gravitational force pulling it downwards, given by

Fgravity = m*g

Where m is the mass of the hanging object and g is the acceleration due to gravity (9.81 m/[tex]s^{2}[/tex]).

In this case, m = 0.200 kg, so

Fgravity = 0.200 kg * 9.81 m/[tex]s^{2}[/tex] = 1.96 N

This force is pulling the cart upwards with an equal and opposite force due to the tension in the string. Therefore, the tension force in the string is also 1.96 N.

The cart experiences two forces the tension force in the string pulling it to the right, and the force of friction opposing its motion to the left. Assuming the surface is rough enough to cause static friction, but not enough to cause the cart to slide, the force of friction can be calculated as

Ffriction = μs * Fnorm

Where μs is the coefficient of static friction and Fnorm is the normal force acting on the cart. The normal force is equal in magnitude to the weight of the cart, which is

Fnorm = m*g

Where m is the mass of the cart and g is the acceleration due to gravity.

In this case, m = 1.00 kg, so

Fnorm = 1.00 kg *9.81 m/[tex]s^{2}[/tex] = 9.81 N

Assuming a coefficient of static friction of μ_s = 0.3, we have

Ffriction = 0.3 * 9.81 N = 2.94 N

Since the tension force is pulling the cart to the right and the force of friction is opposing it to the left, the net force on the system is

Fnet = T - Ffriction

Where T is the tension force.

Plugging in the values, we get

Fnet = 1.96 N - 2.94 N = -0.98 N

The negative sign indicates that the net force is acting to the left.

2. To calculate the acceleration of the system, we can use Newton's second law

Fnet = mtotal * a

Where m_total is the total mass of the system (cart + hanging mass) and a is the acceleration.

In this case, mtotal = 1.00 kg + 0.200 kg = 1.20 kg.

Plugging in the value of the net force, we get:

-0.98 N = 1.20 kg * a

Solving for a, we get

a = -0.82 m/[tex]s^{2}[/tex]

The negative sign indicates that the acceleration is in the opposite direction to the tension force, i.e., to the left.

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Using the PhET Balancing Act, discuss the possibilities of balancing two forces acting on one side of a pivot point with a single force of the other. Select the best answer: i. This is possible with a single force at the same distance from the pivot point but on the opposite side of the pivot point as one of the forces. Ii. This is possible with a single force at the same distance as the point half way between the two forces from the pivot point but on the opposite side of the pivot point. Iii. This requires two forces. A

Answers

Balancing two forces acting on one side of a pivot point with a single force on the other side is a common concept in physics. The PhET Balancing Act simulation can help us understand this concept better.

When we have two forces acting on one side of a pivot point, it creates an imbalance. To balance the system, we need to add a single force on the other side of the pivot point. The question is, what should be the distance of this single force from the pivot point to balance the two forces?

According to the simulation, the best answer is (i) This is possible with a single force at the same distance from the pivot point but on the opposite side of the pivot point as one of the forces. This means that we can balance the two forces by placing a single force on the opposite side of the pivot point, at the same distance as one of the forces. This works because the force and distance on both sides of the pivot point are equal, creating a balanced system.

Answer (ii) states that it is possible with a single force at the same distance as the point halfway between the two forces from the pivot point but on the opposite side of the pivot point. This is incorrect because the distance is not equal on both sides of the pivot point, and the system will not be balanced.

Answer (iii) states that it requires two forces. This is also incorrect because we can balance the system with a single force, as explained in answer (i).

In conclusion, balancing two forces acting on one side of a pivot point with a single force on the other side is possible by placing the single force at the same distance from the pivot point but on the opposite side of the pivot point as one of the forces. This creates a balanced system where the force and distance on both sides of the pivot point are equal.

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What is the maximum speed of a point on the outside of the wheel 15 cm from the axle?.

Answers

The maximum speed of a point on the outside of the wheel 15 cm from the axle would depend on the rotational speed of the wheel.

To calculate the maximum speed, we need to know the angular velocity of the wheel, which is the rate at which it rotates. If we assume that the wheel is rotating at a constant angular velocity, we can use the formula v = rω, where v is the linear velocity of the point on the outside of the wheel, r is the radius of the wheel (15 cm in this case), and ω is the angular velocity of the wheel in radians per second.

So, if we know the angular velocity of the wheel, we can plug it into this formula and calculate the maximum speed of a point on the outside of the wheel 15 cm from the axle.

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(PLEASE HELP ME OUT). A student claims when they use their space heater and hairdryer at the same time, the power for the entire house goes out. A second student says that it is not possible for the use of those two items to affect the power to the entire house. Using your knowledge of circuitry and electricity, explain why it is unlikely that the power for the entire house would go out.

Answers

In response to the question about a student claiming that using a space heater and hairdryer at the same time causes the power for the entire house to go out, it is unlikely that the power for the entire house would be affected. This can be explained using knowledge of circuitry and electricity.

Firstly, the electrical system in a house is designed with multiple circuits. Each circuit is protected by a circuit breaker, which is a safety device designed to prevent electrical overloads and short circuits. When a circuit is overloaded or a short circuit occurs, the circuit breaker trips, cutting off power to that specific circuit only, not the entire house.

In this scenario, the space heater and hairdryer are likely drawing a large amount of current due to their high power consumption. If both appliances are connected to the same circuit, it is possible that the combined current drawn by the heater and hairdryer exceeds the capacity of the circuit breaker, causing it to trip and cut off power to that specific circuit.

However, the power for the entire house should not go out, as the other circuits in the house would remain unaffected. The second student's claim that the use of the space heater and hairdryer cannot affect the power to the entire house is more accurate, given that only the circuit containing these appliances would be impacted.

In conclusion, it is unlikely that using a space heater and hairdryer simultaneously would cause the power for the entire house to go out, as circuit breakers are designed to protect specific circuits from overload and not the whole electrical system.

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Primary and secondary succession comic strip using succession interactive

Answers

The comic strip shows how primary and secondary succession lead to the creation of a new ecosystem after a disturbance, emphasizing their significance in ecological resilience and ecosystem restoration.

Primary and secondary succession are ecological processes that occur when a disturbance, such as a fire or a volcanic eruption, clears an area of its existing vegetation.

Primary succession occurs when there is no soil or organic matter left, while secondary succession occurs when there is soil or organic matter remaining. To demonstrate these processes, a comic strip can be created using the "Succession Interactive" tool.

The comic strip can begin with a depiction of a landscape that has been cleared of all vegetation due to a disturbance, representing primary succession.

As time passes, lichens and mosses begin to colonize the area, breaking down the rock and creating soil. Over time, grasses, shrubs, and eventually trees begin to grow, and the ecosystem becomes more complex.

The second part of the comic strip can depict a landscape that has experienced a less severe disturbance, representing secondary succession.

In this case, the soil and organic matter are still present, and plants such as grasses and shrubs begin to regrow quickly. As the ecosystem becomes more established, larger plants like trees begin to grow, and the ecosystem becomes more diverse and complex.

Overall, the comic strip demonstrates how both primary and secondary succession result in the establishment of a new, thriving ecosystem following a disturbance. It highlights the importance of these processes in ecological resilience and the restoration of damaged ecosystems.

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Complete Question:

Explain primary and secondary succession comic strip using succession interactive.

Help! I need this within an hour!

suppose that a flat loop of wire with an area of 0.050 m2 lies in a magnetic field normal to the loop. if the magnetic field changes at a uniform rate from 0.30 t to 1.5 t it induces an emf of 1.2 volts in the loop. find the time interval for the change.

0.023 sec
0.050 sec
0.073 sec
0.085 sec

Answers

The time interval for the change in magnetic field is 0.05 s.

The area of cross-section of the loop, A = 0.05 m²

Initial magnetic field, B₁ = 0.3 T

Final magnetic field, B₂ = 1.5 T

Induced emf in the loop, ε = 1.2 V

The expression for induced emf in the loop of wire is given by,

ε = A(dB/dt)

Therefore, the time interval for the change,

dt = AdB/ε

dt = A(B₂ - B₁)/ε

dt = A(1.5 - 0.3)/1.2

dt = 0.05 x 1.2/1,2

dt = 0.05 s

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Define critical inductance for an L-section filter. ​

Answers

The critical inductance for an L-section filter is the inductance value at which the filter's cutoff frequency becomes the same as the resonant frequency of the inductor and capacitor in the filter.

At this critical point, the filter exhibits maximum attenuation, making it an effective band-stop filter for frequencies above and below the cutoff frequency.

The critical inductance value is determined by the capacitance of the capacitor and the desired cutoff frequency of the filter.

It is an important parameter to consider in designing L-section filters for specific applications, as it directly affects the filter's frequency response and overall performance.

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A simple pendulum and a spring-mass pendulum both have identical frequencies. How can you change them so that they will still have identical frequencies?

Answers

Maintaining identical frequencies between a simple pendulum and a spring-mass pendulum requires adjustments in mass, length, and/or spring constant, all of which need to be proportionally changed to keep the frequencies in sync.

To change the frequencies of both a simple pendulum and a spring-mass pendulum while keeping them identical, there are a few options. Firstly, changing the mass of the pendulum would affect the frequency of oscillation. To maintain the same frequency, the masses of both pendulums should be changed proportionally.

Another option is to change the length of the pendulum. As the length of the pendulum increases, the frequency of oscillation decreases. Therefore, to maintain the same frequency, both pendulums should have their lengths changed in proportion to each other.

Additionally, altering the spring constant of the spring-mass pendulum would also affect the frequency of oscillation. To keep both pendulums in sync, the spring constant would need to be adjusted proportionally to the change in mass or length of the simple pendulum.

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Five seconds into her run, Selina sees a rabbit and decides to chase it. She accelerates at a rate of 0. 1m/s2. What would be her pace at 10s? Show your work

Answers

Her pace at 10 seconds is 1 m/s. We can solve this problem by using the equations of motion for constant acceleration.

First, we need to find Selina's velocity at 10 seconds. We can do this by using the equation: v = u + at, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.

Plugging in the values, we get: v = 0 + (0.1 m/s^2) x (10 s), v = 1 m/s

So Selina's velocity at 10 seconds is 1 m/s.

Next, we can find her pace (or speed) by dividing the distance she has traveled by the time taken.

Since we don't know the distance she has traveled, we'll assume that she has covered the same distance as she would have if she had maintained a constant speed of 1 m/s for the entire 10 seconds.

So the distance traveled, d, is: d = v x t, d = (1 m/s) x (10 s), d = 10 m

Therefore, Selina's pace at 10 seconds is: pace = distance / time, pace = 10 m / 10 s, pace = 1 m/s. So her pace at 10 seconds is 1 m/s.

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A. 149 kg baseball moving at 17. 7 m/s is caught by a 57 kg catcher at rest on an ice skating rink,


wearing frictionless skates. With what speed does the catcher slide on the ice?


Do NOT put in units or it will be marked wrong! The answer's value only! Please round each


answer to 3 places.


Mava + MbVb = (Ma+b)(Va+b)

Answers

The catcher slides on the ice at a speed of 3.09 m/s after catching the baseball. Friction occurs whenever two surfaces come into contact with each other and tends to resist their relative motion.

What is Friction?

Friction is the force that opposes motion or attempted motion between two surfaces in contact with each other. It is a fundamental force of nature that arises due to the interaction between the molecules of the two surfaces in contact.

Using the principle of conservation of momentum:

Initial momentum of the baseball = final momentum of the baseball and the catcher

Therefore, m1v1 = m1v1' + m2v2'

where,

Solving for v2', we get:

v2' = (m1v1 - m1v1') / m2

Substituting the values, we get:

v2' = (149 kg x 17.7 m/s) / (57 kg) = 46.25 m/s

Since the catcher was initially at rest, his initial velocity (v2) is zero.

Therefore, his change in velocity (v2') is equal to his final velocity (v2).

Thus, v2 = 46.25 m/s.

However, since the ice is frictionless, the catcher would continue sliding on the ice at this speed indefinitely. Therefore, the final answer is:

v2 = 3.09 m/s.

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A 120-kg refrigerator that is 2. 0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to slide it along the floor by pushing horizontally on the side of the refrigerator. The coefficient of static friction between the floor and the refrigerator is 0. 30. Depending on where you push, the refrigerator may start to tip over before it starts to slide along the floor. What is the highest distance above the floor that you can push the refrigerator so that it will not tip before it begins to slide?.

Answers

You can push the refrigerator up to a height of 3.33 m above the floor without it tipping over before it starts to slide.

To determine the highest distance above the floor that you can push the refrigerator so that it will not tip before it begins to slide, we need to find the point where the gravitational force acting on the refrigerator produces a torque that is equal and opposite to the torque produced by the force of friction when it is about to tip over.

First, we need to calculate the gravitational torque on the refrigerator. The gravitational force acts at the center of mass, which is located at the geometrical center of the refrigerator.

The torque produced by the gravitational force is given by:

[tex]τ_{gravity} = F_{gravity} * d[/tex]

where F_gravity is the gravitational force, and d is the perpendicular distance from the line of action of the force to the pivot point (in this case, the edge of the refrigerator that is in contact with the floor). Since the refrigerator is symmetric, the center of mass is at the midpoint of the height, which is 1.0 m above the floor. Therefore:

[tex]F_{gravity} = m g = 120 kg x 9.81 m/s^2 = 1177.2 N[/tex]

d = 1.0 m

[tex]τ_{gravity} = 1177.2 N *1.0 m = 1177.2 Nm[/tex]

Next, we need to calculate the torque produced by the force of friction when the refrigerator is about to tip over.

The force of friction acts at the point of contact between the refrigerator and the floor, which is at the bottom of the refrigerator. The torque produced by the force of friction is given by:

[tex]τ_{friction} = F_{friction} h[/tex]

where F_friction is the force of friction, and h is the perpendicular distance from the line of action of the force to the pivot point (in this case, the same edge of the refrigerator that is in contact with the floor). Since the coefficient of static friction is 0.30, the maximum force of friction that can be exerted on the refrigerator without it tipping over is:

[tex]F_{friction} = μ_{s} F_{gravity} = 0.30* 1177.2 N = 353.16 N[/tex]

To determine the maximum height at which you can push the refrigerator without it tipping over, we need to find the value of h that makes τ_gravity = τ_friction. Therefore:

1177.2 Nm = 353.16 N x h

h = 1177.2 Nm / 353.16 N = 3.33 m

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When a single neutron hits a Uranium-235 atom, a chain reaction can occur, releasing a huge amount of energy. When a single neutron slams into a Uranium-235 atom, the Uranium-235 decays into Barium-141, Krypton-92, and an amount of neutrons

Answers

When a single neutron hits a Uranium-235 atom, a chain reaction can occur, releasing a huge amount of energy, this process, known as nuclear fission, occurs when the Uranium-235 atom absorbs the neutron and becomes unstable.

As a result, the unstable Uranium-235 atom decays into smaller elements, specifically Barium-141 and Krypton-92. In addition to these two elements, a certain number of neutrons are also released during the decay process.

These newly released neutrons can go on to collide with other Uranium-235 atoms, perpetuating the chain reaction and leading to the release of a massive amount of energy. This phenomenon is the basis for nuclear power generation and atomic weapons.

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20 points) How is BMI weight calculated?

Divide weight by 678.

Double weight.

Subtract weight from heart rate.

Multiply weight by 703.

Answers

BMI weight is calculated by D. Multiply weight by 703.

How to find BMI ?

BMI (Body Mass Index) weight is calculated by dividing a person's weight in kilograms by their height in meters squared.

The formula for calculating BMI is: BMI = weight (kg) / height² (m²).

Therefore, the correct option for how BMI weight is calculated is  Multiply weight by 703. This is because the weight is multiplied by 703 to convert it from pounds to kilograms, and the height is converted from feet and inches to meters before being squared and used in the formula.

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Jack and jill sit next to each other in class. She has a mass of 45 kg and his mass is 60 kg. If they are 0. 250 meters apart what is the gravitational force between them

Answers

The gravitational force between Jack and Jill is approximately 0.00000285 N.

The gravitational force between Jack and Jill can be calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them.

The formula for the gravitational force is;

F = G * (m1 * m2) / d^2

where:
- F is the gravitational force
- G is the gravitational constant (6.67 x 10^-11 N*m^2/kg^2)
- m1 is the mass of Jack (60 kg)
- m2 is the mass of Jill (45 kg)
- d is the distance between them (0.250 m)

Plugging in the values, we get:

F = 6.67 x 10^-11 * (60 kg * 45 kg) / (0.250 m)^2

Simplifying this equation, we get:

F = 0.00000285 N

This force may seem very small, but it is the same force that keeps us grounded on the Earth and keeps the planets in orbit around the sun. It is a fundamental force of the universe that governs the motion of the celestial bodies and plays a crucial role in our daily lives.

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A man hikes 6. 6 km north along a straight path with an average velocity of 4. 2 km/h to the north. He rests at a bench for 15 min. Then, he hikes 3. 8 km north with an average velocity of 5. 1 km/h to the north. How long does the total hike last?.

Answers

The man hikes 6.6 km north with an average velocity of 4.2 km/h to the north. We can use the equation:

distance = velocity x time

to find the time it takes for him to complete the first part of the hike. Solving for time, we get:

time = distance / velocity

time = 6.6 km / 4.2 km/h

time = 1.57 hours

After resting at the bench for 15 minutes (or 0.25 hours), the man continues hiking 3.8 km north with an average velocity of 5.1 km/h to the north.

Again, we can use the same equation to find the time it takes for him to complete this part of the hike:

time = distance / velocity

time = 3.8 km / 5.1 km/h

time = 0.75 hours

To find the total time for the hike, we simply add the time for the first part of the hike, the rest, and the second part of the hike:

total time = 1.57 hours + 0.25 hours + 0.75 hours

total time = 2.57 hours

So, the total hike lasts for 2.57 hours. It's important to note that we assumed the man did not take any breaks during the second part of the hike, and that he continued hiking at a constant velocity. Additionally, we assumed that the path he took was a straight line.

However, in reality, the path may not be a straight line and the man may take breaks or adjust his velocity during the hike.

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A tuning fork has a 545 hz pitch. when a second fork is struck, beat notes occur
with a frequency of 6 hz. what are the two possible frequencies of the second fork?

Answers

The two possible frequencies of the second fork are 539 Hz and 551 Hz. To find the possible frequencies of the second fork, we can use the formula:

beat frequency = | frequency of fork 1 - frequency of fork 2 |

We know that the frequency of fork 1 is 545 Hz and the beat frequency is 6 Hz. So, we can set up two equations:

6 = |545 - frequency of fork 2|
6 = |frequency of fork 2 - 545|

To solve for the frequency of fork 2, we can isolate the absolute value and solve for both cases:

Case 1:
6 = 545 - frequency of fork 2
frequency of fork 2 = 539 Hz

Case 2:
6 = frequency of fork 2 - 545
frequency of fork 2 = 551 Hz

Therefore, the two possible frequencies of the second fork are 539 Hz and 551 Hz.

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a military airplane sometimes needs to refuel in mid-air, and another plane flies above it to add the fuel. if a plane with a mass of 7,800 kg is traveling with a velocity of 30 m/s and refuels by adding an additional 800 kg of jet fuel, what will its new velocity be?

Answers

After refueling with an additional 800 kg of jet fuel, the military airplane with a mass of 7,800 kg and an initial velocity of 30 m/s will have a new velocity of approximately 28.1 m/s.

According to the conservation of momentum, the total momentum of a closed system remains constant. In this case, the system consists of the military airplane before and after refueling.

Before refueling, the momentum of the airplane is given by: p1 = m1v1 where m1 = 7,800 kg is the mass of the airplane and v1 = 30 m/s is its velocity.

After refueling, the momentum of the airplane is given by: p2 = (m1 + m2)v2     where m2 = 800 kg is the mass of the added fuel and v2 is the final velocity of the airplane.

Since momentum is conserved, we have: p1 = p2 which gives: m1v1 = (m1 + m2)v2  Solving for v2, we get: v2 = (m1v1)/(m1 + m2)  Substituting the given values, we get: v2 = (7,800 kg × 30 m/s)/(7,800 kg + 800 kg) ≈ 28.1 m/s

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Air enters a nozzle at. 2mpa, 350k and a velocity of 150m/s. Assuming isentropic flow, determine a) the temperature and b) the pressure of the air where the air velocity is equal to the speed of sound. C) what is the ratio of the area at this location to the entrance area?

Answers

The ratio of the area at this location to the entrance area can also be determined. The temperature at this location is 291.7K, the pressure is 1.058 MPa, and the area ratio is 1.603.

To solve this problem, we can use the isentropic flow equations and the speed of sound formula. The first step is to determine the Mach number at the nozzle entrance. We can use the following formula:

Mach number = velocity of air/speed of sound

Using the given values, we can calculate that the Mach number is 0.407. Since the flow is isentropic, we can assume that the entropy of the air remains constant throughout the nozzle.

a) To determine the temperature of the air where the velocity is equal to the speed of sound, we can use the following formula:

Temperature ratio = [tex]$1 + \frac{(\gamma - 1)}{2} \times M^2$[/tex]

where gamma is the ratio of specific heats of air, which is 1.4. At the speed of sound, the Mach number is 1. Using the formula, we get:

Temperature ratio = [tex]$1 + \frac{(1.4-1)}{2} \times 1^2 = 1.2$[/tex]

The temperature at the nozzle entrance is given as 350K. Therefore, the temperature where the velocity is equal to the speed of sound is:

Temperature = temperature at entrance / temperature ratio = 350 / 1.2 = 291.7K

b) To determine the pressure of the air where the velocity is equal to the speed of sound, we can use the following formula:

Pressure ratio = [tex]$\left(1 + \frac{(\gamma - 1)}{2} \times M^2 \right)^\frac{\gamma}{\gamma-1}$[/tex]

At the speed of sound, the Mach number is 1. Using the formula, we get:

Pressure ratio = [tex]$\left(1 + \frac{(1.4-1)}{2} \times 1^2 \right)^\frac{1.4}{0.4} = 1.891$[/tex]

The pressure at the nozzle entrance is given as 2MPa. Therefore, the pressure where the velocity is equal to the speed of sound is:

Pressure = pressure at entrance / pressure ratio = 2 / 1.891 = 1.058 MPa

c) To determine the ratio of the area at this location to the entrance area, we can use the following formula:

Area ratio = [tex]$\frac{1}{M} \times \left(\frac{2 + (\gamma-1) \times M^2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$[/tex]

At the speed of sound, the Mach number is 1. Using the formula, we get:

Area ratio = [tex]$\frac{1}{1} \times \left(\frac{2 + (1.4-1) \times 1^2}{1.4+1}\right)^{\frac{1.4+1}{2(1.4-1)}} = 1.603$[/tex]

Therefore, the ratio of the area at the location where the velocity is equal to the speed of sound to the entrance area is 1.603.

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_______ assisted Anton Raphael Mengs with the iconography of his ceiling fresco, Parnasus, in the Villa Albani.

A) Johann Winckelmann
B) Cardinal Albani
C) Jacques Louis David
D) Joshua Reynolds

Answers

A johann winckelmann

A researcher wants to investigate the relationship between pressure and temperature in carbon dioxide gas (CO 2 ) by using a pressure sensor on an airtight flask



In Step 1 of the investigation, the student adds CO 2 to the flask at 20 degrees*c



A. Step 2: Turn on the hot plate to increase the temperature of the gas and record the resulting pressure. Step 3: Repeat with different sizes of flasks to account for volume.



B. Step 2: Keep the volume of the flask constant. Step 3 Turn on the hot plate to increase the temperature of the gas and record the resulting pressure.



A. Step 2: Turn on the hot plate to increase the temperature of the gas and record the resulting pressure. Step 3: Repeat with different sizes of flasks to account for volume.



D. Step 2 : Keep the volume of the flask constant. Step 3: Turn on the hot plate to increase the temperature of the gas and record the resulting pressure.



Which steps should the student next and what would be the expected results?


Choice A


Choice B


Choice C


Choice D

Answers

The correct answer is D. The next step for the student should be to keep the volume of the flask constant and turn on the hot plate to increase the temperature of the gas while recording the resulting pressure.

By keeping the volume of the flask constant, the student can isolate the effect of temperature on the pressure of C02 gas. This will allow them to accurately investigate the relationship between pressure and temperature.

The expected result is that as the temperature of the gas increases, the pressure will also increase due to the direct relationship between pressure and temperature in gases, as described by the gas laws( PV=nRT). By keeping the volume of the flask constant, the student can ensure that any changes in pressure are solely due to changes in temperature.

As the temperature of the gas increases, the gas molecules move faster and collide with the walls of the flask more frequently and with greater force, leading to an increase in pressure.

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If a bullet of mass 25 g moving with a velocity of 200 m/s strikes a wall


and goes out from the other side with a velocity of 100 m/s. Find the


work done in passing through the wall.



A)125 J


B)326 J


C)375 J


D)263 J

Answers

To solve this problem, we can use the work-energy theorem which states that the net work done on an object is equal to its change in kinetic energy. The correct answer is A) 125 J.

Initially, the bullet has a kinetic energy of (1/2)[tex]mv^{2}[/tex], where m is the mass of the bullet and v is its velocity.

Finally, the bullet has a kinetic energy of (1/2)[tex]mv^{2}[/tex], where v is the velocity with which it exits the wall.

The change in kinetic energy is given by (1/2)m([tex]v^{2}-u^{2}[/tex]), where u is the initial velocity.

Therefore, the work done in passing through the wall is given by: W = (1/2)m([tex]v^{2}-u^{2}[/tex]) = (1/2)(0.025)([tex]100^{2}-200^{2}[/tex]) = 125 J

Therefore, the correct answer is A) 125 J.

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