The angular acceleration of the pulley is approximately [tex]39.22 rad/s^2[/tex].
What does the term "angular acceleration" mean?
The angular acceleration, which is frequently denoted by the symbol and stated in radians per second per second, is the rate at which the angular velocity changes over time.
Here is the calculation:
The net force acting on the 2 kg object is the difference between the tension in the string and the frictional force. Using Newton's second law, we can write:
[tex]F_{net} = ma\\T - f_k = ma[/tex]
The moment of inertia of the pulley can be calculated using the formula for the moment of inertia of a disk:
[tex]I = (1/2)mr^2[/tex]
The torque due to the tension can be calculated as:
[tex]\tau_T = T*(r/2)[/tex]
The torque due to the frictional force can be calculated as:
[tex]\tau_f = f_k*(r/2)[/tex]
The net torque can be calculated as the difference between the torque due to the tension and the torque due to the frictional force:
[tex]\tau_{net} = \tau_T - \tau_f[/tex]
Finally, the angular acceleration can be calculated using Newton's second law for rotational motion:
[tex]\tau_{net} = I*\alpha[/tex]
Substituting the values and solving for α, we get:
[tex]\alpha = (T - f_k)/(1/2mr^2) = (2/3)g(\mu_k - sin\theta)[/tex]
where g is the acceleration due to gravity, [tex]\mu_k[/tex] is the coefficient of kinetic friction, and θ is the angle of the incline.
Using the given values, we get:
[tex]\alpha = (2/3)9.81(0.40 - sin(30)) = 39.22 rad/s^2[/tex]
Therefore, the angular acceleration of the pulley is approximately [tex]39.22 rad/s^2[/tex].
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The period of a simple pendulum of length 1m on a massive planet is 1 sec. What is the acceleration due to gravity on that planet?
The period of a simple pendulum of length 1m on a massive planet is 1 sec. The acceleration due to gravity on that planet is 39.48 m/s^2.
A simple pendulum's period is given by:
T = 2π √(L/g)
Where T is the pendulum's period, L is its length, and g is the acceleration due to gravity.
In this scenario, the pendulum's period is one second and its length is one metre.
So, from above equation, we have:
1 = 2π √(1/g)
Squaring both sides, we get:
1^2 = (2π)^2 (1/g)
Simplifying, we get:
g = (4π^2)/1 = 39.48 m/s^2
Therefore, the acceleration due to gravity on the massive planet is 39.48 m/s^2.
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2. A ball of mass 240 g is moving through the air at 20.0 m/s with a gravitational potential energy of 70 J. With what speed will the ball hit the ground?
The speed at which the ball hit the ground is 31.36 m/s.
What is speed?Speed is the rate of change of distance.
To calculate the speed at which the ball hit the ground, we use the formula below
Formula:
v² = u²+2E/m ................ Equation 1Where:
v = Final speed of the ballu = Initial speed of the ballE = Energym = Mass of the ballFrom the question,
Given:
u = 20 m/sm = 240 g = 0.24 kgE = 70 JSubstituite these values into equation 1
v² = 20²+(2×70)/0.24v² = 400+583.333v² = 400+583.333√v² = √983.33v = 31.36 m/sLearn more about speed here: https://brainly.com/question/29110645
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Sammy Hagar is doing a concert on a stage that travels down the highway at 32 m/s. During warm-
up the band realizes that their concert F needs to be adjusted to sound right to the audience which
is standing still. If a concert Fis 540 Hz, what frequency should they play to make it sound right
To make the concert F sound right to the audience, Sammy Hagar and the band should play the note at a frequency of 607 Hz.
The frequency that the audience will hear, denoted as f', is related to the frequency of the source, f, by the formula: f' = f (v + u) / (v - u)
where v is the speed of sound, u is the speed of the observer relative to the medium, and in this case, v = 343 m/s and u = -32 m/s.
When the stage is moving toward the audience, the relative speed of the sound waves is increased, so the frequency heard by the audience is higher. Using the above formula: f' = 540 Hz (343 + 32) / (343 - 32) = 607 Hz
Therefore, to make the concert F sound right to the audience, Sammy Hagar and the band should play the note at a frequency of 607 Hz.
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A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450
n/m, undergoes shm with an amplitude of 0.040 m. compute (a) the maximum speed
of the glider; (b) the speed of the glider when it is at x = -0.015 m; (c) the magnitude of
the maximum acceleration of the glider; (d) the acceleration of the glider at x = -0.015
m; (e) the total mechanical energy of the glider at any point in its motion
The motion of a 0.500-kg glider attached to an ideal spring with a force constant of k=450m can be analyzed in terms of mechanical energy. Mechanical energy is the sum of kinetic energy and potential energy, and is conserved in a closed system with no external forces acting on it.
As the glider moves back and forth on the spring, its kinetic energy varies with its speed and its potential energy varies with its position. At any point in its motion, the total mechanical energy of the glider is equal to the sum of its kinetic and potential energy.
At the maximum compression of the spring, the glider has zero velocity and maximum potential energy. As it moves away from this point, the spring begins to expand and the glider begins to move faster, converting potential energy into kinetic energy. At the point where the spring is fully extended, the glider has maximum velocity and zero potential energy.
As the glider continues to move back towards the spring's rest position, it begins to slow down and convert kinetic energy back into potential energy. At the point of maximum compression again, the glider has zero velocity and maximum potential energy once more.
Throughout its motion, the total mechanical energy of the glider remains constant, as there are no external forces acting on the system. This means that the sum of the kinetic and potential energy at any point in its motion is equal to the total mechanical energy of the system.
In summary, the mechanical energy of a glider attached to an ideal spring can be analyzed at any point in its motion by considering the conversion of potential energy into kinetic energy and vice versa. The total mechanical energy of the system is constant throughout its motion, making it a useful tool for analyzing the behavior of the glider on the spring.
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lighting flashes and you hear a thunder clap 4 seconds later. the velocity of sound is 340 m/s. how far away did the lighting strike
Answer:
The lightning struck 1,360 meters away
Explanation:
1. List knowns
Speed of sound = 340 m/s
Time = 4 s
2. Find formula that uses above knowns
Speed = Distance / Time
Distance = Time x Speed
3. Substitute
Distance = 4 s x 340 m/s
Distance = 1360 meters
At a particular instant in the flight the glider is losing 1. 00 m of vertical height for every 6. 00 m that it goes forward horizontally. At [3 marks] this instant, the horizontal speed of the glider is 12. 5 m s. Calculate the velocity of the glider. Give your answer to an appropriate number of significant figures
12.7 ms−1 is the velocity of the glider
Define velocity.
When an object is moving, its velocity is the rate at which it is changing position as seen from a specific point of view and as measured by a specific unit of time.
Velocity can be defined as the rate at which something moves in a specific direction. as the speed of a car driving north on a highway or the speed at which a rocket takes off.
Given,
Horizontal speed ∣→vh∣ =12.5 ms−1
t=Distance/Speed ⇒
t=6.00/12.5= 0.48s
Vertical speed
∣→vv∣=1.00/0.48=2.083 ms−1
∣→v∣=√(12.52)+(2.083)2
= 12.7 ms−1
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What type of radioactive decay is this process? An example of? 14 6c 1417n +0 negative one negative plus the v
The type of radioactive decay of carbon to nitrogen is beta-minus decay.
A kind of radioactive decay called beta-minus involves the emission of electrons and antineutrinos from the nucleus as well as the transformation of neutrons into protons, which raises the atomic number of the atom..
This increases the atomic number of the nucleus by one and leaves the mass number unchanged. The question mentions the decay of carbon-14 (C) to nitrogen-14 (N) as an example of beta-minus decay in the given reaction.
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Complete question - What type of radioactive decay is this process? An example of?
¹⁴C → ¹⁴N + e⁻ + v
What is the approximate electrostatic force between two protons each having a charge of +1. 6 x 10-19 C separated by a distance of 1. 0 × 10–6 meter?
A)
2. 3 × 10–16 N and repulsive
B)
2. 3 × 10–16 N and attractive
C)
9. 0 × 1021 N and repulsive
D)
9. 0 × 1021 N and attractive
The approximate electrostatic force between two protons each having a charge of +1. 6 x 10-19 C separated by a distance of 1. 0 × 10–6 meter A) 2.3 × 10^–16 N and repulsive.
To calculate the electrostatic force between two protons, we can use Coulomb's Law:
F = (k * q1 * q2) / r^2
where F is the electrostatic force, k is Coulomb's constant (8.99 × 10^9 N m^2 C^−2), q1 and q2 are the charges of the two protons, and r is the distance between them.
Given: q1 = q2 = +1.6 × 10^-19 C, r = 1.0 × 10^-6 m
Now, plug the values into the formula:
F = (8.99 × 10^9 N m^2 C^−2 * (1.6 × 10^-19 C)^2) / (1.0 × 10^-6 m)^2
F ≈ 2.3 × 10^-16 N
Since both charges are positive, the electrostatic force will be repulsive. Therefore, the correct answer is:
A) 2.3 × 10^–16 N and repulsive
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During super bowl weekend, the NFL sets up a receiver on a stationary hovercraft. A
. 257 kg football is thrown at 9. 76 m/s to a receiver and hovercraft with a total mass of
98. 6 kg. When the ball is caught what is the new speed of the system?
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)
The new speed of the system when the ball is caught is approximately 0.025 m/s
To solve this problem, we will use the conservation of momentum equation:
MaVa + MbVb = (Ma + Mb)(Va+b)
where Ma is the mass of the football (0.257 kg), Va is the velocity of the football (9.76 m/s), Mb is the mass of the receiver and hovercraft (98.6 kg), and Vb is the initial velocity of the receiver and hovercraft (0 m/s, since it is stationary).
0.257 kg * 9.76 m/s + 98.6 kg * 0 m/s = (0.257 kg + 98.6 kg) * (Va+b)
2.50632 kg*m/s = 98.857 kg * (Va+b)
Now, we will solve for Va+b:
Va+b = 2.50632 kg*m/s / 98.857 kg
Va+b ≈ 0.025 m/s
So, the new speed of the system when the ball is caught is approximately 0.025 m/s, rounded to three decimal places.
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A conducting coil of 2250 turns is connected to a galvanometer, and the total resistance of the circuit is 30 ω. the area of each turn is 5. 00 × 10-4 m2. this coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. the amount of charge that is inducedto flow around the circuit is measu
When a conducting coil is moved into a region with a magnetic field, an electromotive force (EMF) is induced in the coil, which causes a current to flow through the circuit.
The magnitude of the induced EMF can be calculated using Faraday's law of electromagnetic induction, which states that the induced EMF is equal to the rate of change of magnetic flux through the coil.
In this case, the coil has 2250 turns and an area of 5.00 × 10^-4 m^2 per turn. If the coil is moved into a region with a magnetic field, the magnetic flux through the coil will change, inducing an EMF in the circuit.
Assuming that the normal to the coil is parallel to the magnetic field, the magnitude of the induced EMF can be calculated as follows:
EMF = -N(dΦ/dt)
where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux through the coil.
The magnetic flux through the coil is given by:
Φ = BA
where B is the magnetic field strength and A is the area of the coil.
Assuming that the magnetic field is uniform and perpendicular to the coil, the magnetic flux through the coil can be written as:
Φ = BNA
The rate of change of magnetic flux through the coil is given by:
dΦ/dt = BNA(v/A) = BNV
where v is the velocity of the coil.
Substituting the values given, we get:
EMF = -2250(5.00 × 10^-4 m^2)(B)(V)/30 Ω
The negative sign indicates that the direction of the induced EMF is opposite to the change in magnetic flux.
The amount of charge that flows around the circuit can be calculated using the equation:
Q = EMF/R
where R is the total resistance of the circuit.
Substituting the values given, we get:
Q = (-2250)(5.00 × 10^-4 m^2)(B)(V)/(30 Ω)
Therefore, the amount of charge induced to flow around the circuit depends on the strength of the magnetic field, the velocity of the coil, and the resistance of the circuit.
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Bob has been asked to produce a fuse that has 3. 2 ohms of resistance. He cannot change he thickness of the wire but can change its length. Explain in detail how bob could work out what length of wire to use. You will need to describe the experiment he will need to carry out , any hazards and any variables involved
The wire's resistance at different lengths and analyzing the data, Bob can determine the appropriate length of wire needed to achieve a resistance of 3.2 ohms.
To determine the length of wire Bob needs to achieve a resistance of 3.2 ohms, he can perform an experiment using the wire to measure its resistance at different lengths. Here's a step-by-step explanation of how Bob can carry out the experiment:
1. Gather materials: Bob will need the wire, a power supply (e.g., a battery), an ammeter (to measure current), and a voltmeter (to measure voltage). Ensure all equipment is properly calibrated and suitable for the current and voltage levels.
2. Design a circuit: Bob should set up a simple circuit consisting of the power supply connected in series with the wire, the ammeter to measure the current passing through the wire, and the voltmeter connected across the wire to measure the voltage drop.
3. Safety precautions: It is important for Bob to follow safety protocols while conducting the experiment. He should handle the wire and electrical equipment with care, avoid touching exposed wires, and ensure the circuit is properly insulated. Additionally, he should wear appropriate safety gear such as gloves and goggles.
4. Initial wire length: Bob should start with an initial length of wire and measure its resistance using a multimeter or an ohmmeter. This measurement will serve as the baseline value.
5. Adjusting wire length: Bob can then modify the length of the wire by cutting or extending it. For each length, he needs to ensure the wire is securely connected in the circuit.
6. Recording data: At each wire length, Bob should record the current (I) and voltage (V) values from the ammeter and voltmeter, respectively. These readings will help him calculate the resistance using Ohm's law: R = V/I.
7. Repeat measurements: Bob should repeat the measurements for several different wire lengths to gather enough data points to analyze and determine a trend.
8. Data analysis: Bob can plot a graph of wire length (x-axis) against resistance (y-axis) using the recorded data. By observing the relationship between wire length and resistance, he can identify the length of wire that corresponds to a resistance of 3.2 ohms.
Variables and Hazards:
Independent variable: Wire length. Bob can manipulate this variable by changing the wire's length.
Dependent variable: Resistance. Bob will measure this variable and use it to determine the relationship with the wire length.
Control variables: Bob should keep other factors constant throughout the experiment, such as the thickness of the wire and the material used.
Hazards: The main hazards involved in this experiment are electrical hazards, including electric shock and short circuits. Bob should ensure the circuit is properly insulated, handle the wires and equipment safely, and follow electrical safety guidelines.
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A honey bee's wings beat at 230 beats
per second. If the speed of sound in air
is 340 m/s, what is the wavelength of the
sound wave?
1 pt: knowns/unknown
1 pt: write the equation
1 pt: solve
1 pt: correct answer (you can round to
one decimal place)
please answer right away
The wavelength of the sound wave is approximately 1.5 meters.
1 pt: Knowns/Unknown:
- Frequency (f) = 230 beats per second (Hz)
- Speed of sound (v) = 340 m/s
- Wavelength (λ) = Unknown
1 pt: Write the equation:
The equation relating the speed of sound, frequency, and wavelength is: v = f * λ
1 pt: Solve:
To find the wavelength (λ), rearrange the equation: λ = v / f
1 pt: Correct answer (rounded to one decimal place):
λ = 340 m/s / 230 Hz ≈ 1.5 m
The wavelength of the sound wave is approximately 1.5 meters.
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An individual receives frequent injections of drugs, which are administered in a small examination room at a clinic. The drug itself causes increased heart rate but after several trips to the clinic, simply being in a small room causes an increased heart rate
The repeated association of the drug injection with the small examination room has led to classical conditioning, resulting in an increased heart rate response to just being in the room.
This phenomenon can be explained through classical conditioning. Classical conditioning is a type of learning in which an organism learns to associate two stimuli together, resulting in a change in behavior.
In this case, the drug injection is the unconditioned stimulus (UCS) that naturally elicits an unconditioned response (UCR) of an increased heart rate.
However, over time, the small examination room has become a conditioned stimulus (CS) that has been associated with the drug injection and now elicits a conditioned response (CR) of an increased heart rate. This means that just the sight or thought of the examination room triggers the same physiological response as the drug itself.
This type of classical conditioning can have both positive and negative effects. On one hand, it can be beneficial for patients who are receiving treatment, as it can help them to anticipate and prepare for the effects of the drug.
On the other hand, it can also lead to a heightened anxiety or fear response in patients who may associate the examination room with negative experiences.
In summary, the repeated association of the drug injection with the small examination room has led to classical conditioning, resulting in an increased heart rate response to just being in the room.
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The force of attraction that a -40. 0 μc point charge exerts on a 108 μc point charge has magnitude 4. 00 n. How far apart are these two charges? (k = 1/4πε0 = 8. 99 × 109 n ∙ m2/c2) show your work
The force of attraction that a -40. 0 μc point charge exerts on a 108 μc point charge has magnitude 4 N. then these two charges are apart by the distance 3.11 m.
According to the law, the strength of the electrostatic force of attraction or repulsion between two point charges is inversely proportional to the square of the distance between them and directly proportional to the product of the magnitudes of the charges. Coulomb investigated the repellent force between things with identical electrical charges:
Given,
q₁ = 40 × 10⁻⁶ C
q₂ = 108 × 10⁻⁶ C
F = 4 N
1/4πε0 = 8. 99 × 10⁹
Coulomb's law is given by,
F = q₁q₂ ÷ 4π∈r²
4 N = - 40 × 10⁻⁶ C × 108 × 10⁻⁶ C × 8. 99 × 10⁹ ÷ r²
4 N = 38.83÷ r²
r² = 38.83÷ 4
r² = 9.7
r = 3.11 m
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Rolanda sees an error in her friend’s graphic organizer comparing electrical and gravitational forces. a venn diagram with two intersecting circles. the circle on the left is labeled gravitational force. the circle on the right is labeled electrical force. there is an x in the circle on the left with infinite reach and depends on mass. in the circle on the right is z with depends on charge. there is a y in the intersecting area. which change should rolanda suggest to her friend to correct the error? the note about mass belongs in region z, and the note about charge belongs in region x. the note about mass belongs in region y, and the note about infinite reach belongs in region z. the note about charge belongs in region y. the note about infinite reach belongs in region y.
Rolanda needs to advise her friend to place the note about mass in region z and the note about charge in region x to fix the mistake in the Venn diagram. Therefore, the correct answer is option A.
Rolanda should suggest to her friend that the note about mass belongs in region z, and the note about charge belongs in region x. This correction is necessary because gravitational force depends on mass, while electrical force depends on charge.
The x in the circle on the left with infinite reach and depends on mass is incorrect because gravitational force does not have infinite reach. It only acts between objects with mass that are in close proximity to each other. The y in the intersecting area is also incorrect because there is no force that is common to both gravitational and electrical forces.
On the other hand, the z in the circle on the right with depends on charge is correct because electrical force depends on the charge of the objects involved. By suggesting that the note about mass belongs in region z and the note about charge belongs in region x, the Venn diagram will accurately represent the differences between gravitational and electrical forces.
In summary, Rolanda should suggest to her friend that the note about mass belongs in region z and the note about charge belongs in region x. Therefore, the correct answer is option A.
This will correct the error in the Venn diagram and accurately represent the differences between gravitational and electrical forces. It is important to understand these differences in order to properly understand the behavior of objects in our world.
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The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers. When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.
A)At what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60∘ above the water surface. Express your answer in meters per second.
Answer:
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them.
Explanation:
A 2000-kg ferris wheel accelerates from rest to an angular speed of 20 rad/s in 12 s. approximate the ferris wheel as a circular disk with a radius of 30 m. what is the net torque on the wheel
The net torque on the Ferris wheel is approximately 1,500,000 N*m.
To find the net torque on the Ferris wheel, we'll need to use the following formula: τ = I * α, where τ represents torque, I is the moment of inertia, and α is the angular acceleration.
First, we need to find the angular acceleration (α). Since the Ferris wheel accelerates from rest (initial angular speed = 0) to an angular speed of 20 rad/s in 12 s, we can use the formula: α = (final angular speed - initial angular speed) / time.
α = (20 rad/s - 0 rad/s) / 12 s = 20/12 rad/s² = 5/3 rad/s²
Next, we'll find the moment of inertia (I) for a circular disk with a mass (m) of 2000 kg and a radius (r) of 30 m: I = (1/2) * m * r².
I = (1/2) * 2000 kg * (30 m)² = 1000 kg * 900 m² = 900,000 kg*m²
Now, we can find the net torque (τ) using the formula: τ = I * α.
τ = 900,000 kg*m² * 5/3 rad/s² ≈ 1,500,000 N*m
So, the net torque on the Ferris wheel is approximately 1,500,000 N*m.
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an inductor must be selected for a circuit that will exactly match the reactance of a 711.3 nf capacitor in a 120 v, 58.0 hz source. determine the required inductance. g
If an inductor must be selected for a circuit that will exactly match the reactance of a 711.3 nf capacitor in a 120 v, 58.0 hz source, the required inductance for the circuit is 65.0 millihenries.
To determine the required inductance for a circuit that matches the reactance of a 711.3 nf capacitor in a 120 V, 58.0 Hz source, we need to use the formula for calculating reactance.
Reactance is the opposition that an inductor or capacitor offers to alternating current, and it is measured in ohms. The reactance of an inductor is given by the formula X₁ = 2πfL, where X₁ is the inductive reactance in ohms, f is the frequency in Hertz, and L is the inductance in Henrys.
The reactance of a capacitor is given by the formula X₂ = 1/(2πfC), where X₂ is the capacitive reactance in ohms, f is the frequency in Hertz, and C is the capacitance in farads.
To match the reactance of the capacitor, we need to calculate the inductance required to cancel out the capacitive reactance. Therefore, we need to set X₁ equal to X₂ and solve for L.
X₁ = X₂
2πfL = 1/(2πfC)
L = 1/(4π^2f^2C)
Substituting the given values, we get:
L = 1/(4π^2(58.0 Hz)^2(711.3 nF))
L = 65.0 mH
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PLS ANSWER ASAP
1. A gardener uses a wheelbarrow to move 20 kilograms of soil from a compost pile to a flower bed, a distance of 53 meters. The wheelbarrow has a mass of 17 kilograms. he expands 94.5 Newtons of forced. How much work does the gardener do? (1 point)
O 135.4 N
O 5,008.5 N
O 1,961 N
O 3,496 N
2. A force of 30 N is applied to a ball, and it takes the balls 1.5 seconds to travel 4 meters. What is the work done on the ball? (1 point)
O 45 J
O 80 J
O 120 J
O 180 J
3. A mechanic wants to use a compound pulley to lift a go-kart from the ground to work table, a distance of 1.2 meters. Without the pulley, 1,620 Newtons of force would be needed to lift the go-kart. If the pulley has a mechanical advantage of 4, how much force must the mechanic expend? (1 point)
O 1,616 N
O 405 N
O 5,400 N
O 1,350 N
The gardener does a work of 5,008.5 J
The work done is 120 J
The force exerted is 405 N
Work is defined as the product of the force applied to an object and the distance it moves in the direction of that force. In other words, work is done when a force causes an object to move in the same direction as the force. The formula for work is W = F x d,
Given that;
W = Fd
F = force
d = distance
W = work done
Thus;
W = 94.5 N * 53 m
= 5,008.5 J
2) W = Fd
W = 30 N * 4 m
= 120 J
3) MA = 1620/x
x = 1620/4
x = 405 N
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You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeoff of at least 27. 8m/s and can accelerate at 2. 00m/s^2. (a) If the runway is 150m long, can this airplane reach the required speed for takeoff? (b) If not , what minimum length must the runway have?
The minimum runway length required for this airplane to reach the required speed for takeoff is 193.41 meters.
(a) To determine if the airplane can reach the required speed for takeoff on a 150m long runway, we can use the equation: v^2 = u^2 + 2as. Here, v is the final speed (27.8 m/s), u is the initial speed (0 m/s, assuming the plane starts from rest), a is the acceleration (2.00 m/s^2), and s is the distance (150m).
27.8^2 = 0^2 + 2(2.00)(150)
773.64 = 600
Since 773.64 > 600, this airplane cannot reach the required speed for takeoff on a 150m long runway.
(b) To find the minimum runway length required for this airplane to take off, we can rearrange the equation: s = (v^2 - u^2) / 2a.
s = (27.8^2 - 0^2) / (2 * 2.00)
s = 773.64 / 4
s = 193.41m
So, the minimum runway length required for this airplane to reach the required speed for takeoff is 193.41 meters.
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Ms. sison is riding his bike and uses 600 joules of energy per minute. if the bike only does 550 joules of work, how efficient is the bike in percent?
The efficiency of the bike can be calculated by dividing the work output by the energy input and multiplying the result by 100%. In this case, the bike is 91.67% efficient.
The efficiency of a machine is defined as the ratio of the work output to the energy input. In this case, the energy input is given as 600 joules per minute, and the work output is 550 joules.
Therefore, the efficiency of the bike can be calculated using the following formula:
Efficiency = (Work output / Energy input) x 100%
Substituting the given values, we get:
Efficiency = (550 / 600) x 100%
Efficiency = 0.9167 x 100%
Efficiency = 91.67%
This means that the bike is 91.67% efficient, which is the percentage of the energy input that is converted into useful work output. The remaining energy is lost as heat due to friction, air resistance, and other factors.
Therefore, the efficiency of the bike can be improved by reducing these losses through proper maintenance and adjustments.
In summary, the efficiency of the bike can be calculated by dividing the work output by the energy input and multiplying the result by 100%. In this case, the bike is 91.67% efficient.
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an airplane is flying at an elevation of 5150 feet, directly above a straight highway. two motorists are driving cars on the highway on opposite sides of the plane and the angle of depression to one car is 35 degrees and to the other car is 52 degrees, how far apart are the cars?
Answer:
7086.9 feet.
Explanation:
We can see that the two triangles formed by the plane and the cars are similar, because they share a common angle (90 degrees) and have corresponding angles that are equal (the angles of depression). Therefore, we can use the proportionality of corresponding sides to find the distance between the cars. Let x be the distance from the plane to the car with 35 degrees angle of depression, and y be the distance from the plane to the car with 52 degrees angle of depression. Then we have:
x / sin(35) = y / sin(52) = 5150 / sin(90)Cross-multiplying and solving for x and y, we get:x = 5150 x sin(35) / sin(90) x = 2957.8 feety = 5150 x sin(52) / sin(90) y = 4129.1 feetThe distance between the cars is the sum of x and y:d = x + y d = 2957.8 + 4129.1 d = 7086.9 feetThe answer is 7086.9 feet.
A 4.0 kg mass is 1.0 m away from a 7.0 kg mass. What is the gravitational force between the two masses? (Remember to use the gravitational constant, G = 6.67 x 10-11 N x m2/ kg2, in your calculation.)
6.67 x 10 -11 N
1.9 x 10 -9N
6.67 x 10 10N
3.8 N
The gravitational force between the two masses is approximately 1.96 x 10⁻⁹ N. Option B is correct.
The gravitational force between two masses can be calculated using the formula;
F=G x (m₁ x m₂) / r²
Where F is the gravitational force, G is the gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between their centers of mass.
In this case, m₁ = 4.0 kg, m₂ = 7.0 kg, r = 1.0 m, and G = 6.67 x 10⁻¹¹ N x m²/kg². Plugging these values into the formula gives;
F = (6.67 x 10⁻¹¹ N x m²/kg²) x (4.0 kg x 7.0 kg) / (1.0 m)²
F = 1.96 x 10⁻⁹ N
Therefore, the gravitational force is 1.96 x 10⁻⁹ N.
Hence, B. is the correct option.
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--The given question is incomplete, the complete question si
"A 4.0 kg mass is 1.0 m away from a 7.0 kg mass. What is the gravitational force between the two masses? (Remember to use the gravitational constant, G = 6.67 x 10-11 N x m2/ kg2, in your calculation.) A) 6.67 x 10 -11 N B) 1.9 x 10 -9N C) 6.67 x 10 10N D) 3.8 N."--
A pen contains a spring with a constant of 216 N/m. When the tip of the pen is in its retracted position, the spring is compressed 4.10 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.10 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.
Answer:The spring force is conservative, so the work done by the spring force is equal to the negative of the potential energy stored in the spring:
U = -1/2 k x^2
where k is the spring constant and x is the displacement from the unstrained length.
The initial compression of the spring is 4.10 mm = 0.00410 m, and the additional compression is 6.10 mm = 0.00610 m. The total compression of the spring is therefore x = 0.00410 m + 0.00610 m = 0.0102 m.
The potential energy stored in the spring when it is compressed by a distance x is:
U = -1/2 k x^2
Substituting the given values, we get:
U = -1/2 (216 N/m) (0.0102 m)^2
U = -0.0112 J
The work done by the spring force to ready the pen for writing is equal to the change in potential energy:
W = U_final - U_initial
where U_initial is the potential energy of the spring when it is compressed 4.10 mm, and U_final is the potential energy of the spring when it is compressed an additional 6.10 mm.
U_initial = -1/2 (216 N/m) (0.00410 m)^2 = -0.000090 J
U_final = -1/2 (216 N/m) (0.0102 m)^2 = -0.0112 J
W = U_final - U_initial
W = (-0.0112 J) - (-0.000090 J)
W = -0.0111 J
The negative sign indicates that the work done by the spring force is done on the pen (i.e. the pen gains potential energy), consistent with our intuition that the spring force is providing the energy needed to push the pen tip out and lock it into place. Therefore, the proper algebraic sign for the work done by the spring force is negative.
Explanation:
A laser beam is aimed through a circular aperture of diameter 1 mm.
a. If the laser beam is red with a wavelength of 632. 8 nm, what is the angle from the center of the Airy disk to the first dark ring? (2 points)
sin(p) = 632. 8*10^-9 /. 001
sin^-1(. 0006328) =. 0363 degrees
b. If the screen you are projecting the Airy disk onto is 2 m from the aperture, what is the distance between the center of the disk and the first dark ring? (2 points)
Thanks everyone who can help!
The angle from the center of the Airy disk to the first dark ring is 0.0363 degrees, and the distance between the center of the disk and the first dark ring on a screen 2 meters away from the aperture is: 1.268 mm.
a. To find the angle from the center of the Airy disk to the first dark ring, we will use the formula sin(p) = (wavelength) / (aperture diameter). Plugging in the values, we get sin(p) = 632.8 * 10^-9 / 0.001. Then, we calculate the inverse sine, sin^-1(0.0006328) = 0.0363 degrees.
b. To determine the distance between the center of the Airy disk and the first dark ring on a screen that is 2 meters from the aperture, we will use the formula distance = (angle) * (distance to screen).
In this case, distance = 0.0363 degrees * 2 meters.
First, convert the angle to radians: 0.0363 degrees * (pi / 180) = 0.000634 radians.
Then, multiply by the distance to the screen: 0.000634 radians * 2 meters = 0.001268 meters or 1.268 mm.
In summary, the angle from the center of the Airy disk to the first dark ring is 0.0363 degrees, and the distance between the center of the disk and the first dark ring on a screen 2 meters away from the aperture is 1.268 mm.
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Complete question:
A laser beam is aimed through a circular aperture of diameter 1 mm.
a. If the laser beam is red with a wavelength of 632. 8 nm, what is the angle from the center of the Airy disk to the first dark ring? (2 points)
sin(p) = 632. 8*10^-9 /. 001
sin^-1(. 0006328) =. 0363 degrees
b. If the screen you are projecting the Airy disk onto is 2 m from the aperture, what is the distance between the center of the disk and the first dark ring? (2 points)
Write an equation for the strength of the electrostatic force for two charges that are separated by 10 meters. Use the Gizmo to check your equation
F = k * ([tex]q_1[/tex][tex]q_2[/tex])/[tex]r^{2}[/tex], where F is the strength of the electrostatic force, k is Coulomb's constant (9X[tex]10^{9}[/tex] N*[tex]m^{2}[/tex]/[tex]C^{2}[/tex]), [tex]q_1[/tex] and [tex]q_2[/tex] are the charges, and r is the distance between the charges in meters.
What is Magnetic Force?
Magnetic force is a fundamental force of nature that is exerted between moving charged particles, such as electrons or between a magnetic field and a moving charge. This force can cause a magnetic material to experience a force of attraction or repulsion depending on its orientation with respect to the magnetic field.
Using the Gizmo to check, we can input two charges and a distance of 10 meters to calculate the electrostatic force between them. The result should match the output of the equation F = k * ([tex]q_1[/tex]*[tex]q_2[/tex])/[tex]r^{2}[/tex].
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How does the ISS maintain equilibrium when there are unbalanced forces?
The ISS maintains equilibrium using thrusters, gyroscopes, and aerodynamically stable components.
The International Space Station (ISS) maintains equilibrium in spite of unbalanced forces through the use of thrusters and gyroscopes. The thrusters can be fired in short bursts to adjust the station's position, while gyroscopes help to maintain its orientation in space.
The station's position and orientation are constantly monitored by ground control, which can send commands to adjust the thrusters and gyroscopes as needed to make equilibrium.
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Suppose that water waves coming into a dock have a velocity of 1.2 m/s and a wavelength of 2.4 m. with what frequency do these waves meet the dock
The frequency with which these waves meet the dock is 0.5 Hz.
To calculate the frequency of the water waves meeting the dock, you can use the formula:
Frequency (f) = Velocity (v) / Wavelength (λ)
Given that the velocity (v) is 1.2 m/s and the wavelength (λ) is 2.4 m, you can plug in these values into the formula:
f = 1.2 m/s / 2.4 m
f = 0.5 Hz
So, the frequency with which these waves meet the dock is 0.5 Hz.
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Suppose you can jump 1 m on earth. how high would you be able to jump on
mars? the mass of mars is 6.4181023kg and its radius is 3.38106m.
To determine how high you would be able to jump on Mars compared to Earth, we can use the concept of gravitational potential energy.
On both Earth and Mars, the gravitational potential energy (PE) is given by the equation:
PE = mgh
where m is your mass, g is the acceleration due to gravity, and h is the height.
The acceleration due to gravity can be calculated using Newton's law of universal gravitation:
g = (G * M) / (r^2)
where G is the gravitational constant, M is the mass of the celestial body (in this case, Mars), and r is the distance from the center of the celestial body to the surface (in this case, the radius of Mars).
Let's assume your mass is the same on both Earth and Mars.
On Earth, the acceleration due to gravity (g_Earth) is approximately 9.8 m/s^2, and the height (h_Earth) is 1 m.
PE_Earth = m * g_Earth * h_Earth
On Mars, we need to calculate the acceleration due to gravity (g_Mars) using the given mass and radius of Mars. The gravitational constant (G) is approximately 6.67430 × 10^(-11) m^3/(kg s^2).
g_Mars = (G * M) / (r^2)
g_Mars = (6.67430 × 10^(-11) m^3/(kg s^2) * 6.418 * 10^(23) kg) / (3.38106 m)^2
Now we can calculate the height (h_Mars) you would be able to jump on Mars:
PE_Mars = m * g_Mars * h_Mars
Since we are assuming the same mass on both Earth and Mars, we can set the potential energies on Earth and Mars equal to each other and solve for h_Mars:
m * g_Earth * h_Earth = m * g_Mars * h_Mars
h_Mars = (g_Earth / g_Mars) * h_Earth
Now we can substitute the values and calculate the height you would be able to jump on Mars:
h_Mars = (9.8 m/s^2 / g_Mars) * 1 m
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31. Two parallel rails with negligible resistance are 10. 0 cm apart and are connected by a 5. 00 ohm resistor. The circuit also contains two metal rods having resistances of 10. 0 ohm and 15. 0 ohm sliding along the rails (Fig. P31. 31). The rods are pulled away from the resistor at constant speeds of 4. 00 m/s and 2. 00 m/s, respectively. A uniform magnetic field of magnitude 0. 0100 T is applied perpendicular to the plane of the rails. Determine the current in the 5. 00 ohm resistor.
question taken from physics for scientists and engineers, 6th edition. Chapter 31, q. 31
The current in the 5.00 ohm resistor is 0.052 A.
The induced emf in each metal rod is given by e = Blv, where B is the magnetic field strength, l is the length of the metal rod moving in the magnetic field, and v is the velocity of the rod. For the 10.0 ohm rod, the induced emf is e = (0.0100 T)(0.100 m)(4.00 m/s) = 0.00400 V. The current through the 10.0 ohm rod is then I1 = e/R1 = 0.000400 A.
For the 15.0 ohm rod, the induced emf is e = (0.0100 T)(0.100 m)(2.00 m/s) = 0.00200 V. The current through the 15.0 ohm rod is then I2 = e/R2 = 0.000133 A. Since the two rods are connected in series, the current through the 5.00 ohm resistor is the same as the current through the two rods: I = I1 + I2 = 0.000533 A.
Using Ohm's law, the voltage drop across the 5.00 ohm resistor is V = IR = (0.000533 A)(5.00 ohm) = 0.00266 V. Therefore, the current in the 5.00 ohm resistor is I = V/R = (0.00266 V)/(5.00 ohm) = 0.052 A.
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