What is the work applied to a 1000kg car when it is pushed 1000m up along a 10- degree incline at a constant velocity. Assume the coefficient of friction between the car and the ramp is 0.30.

Answer:

Explanation:

56) A

57) E

Answer:

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Answer:

well, the opposed friction is caused because of the weight of the crate,so if the weight is been added more and becomes heavier the friction between the floor and the crate will increase

Friction is the one that opposes motion. Increase in the weight of the body causes an increase friction offered to the relative motion of the surfaces in contact. It is clear that the frictional force will change.

There are several factors that can affect friction either increase or decrease the resistance.

Some of them are:

The friction is directly proportional to the weight of the body on a horizontal surface and hence an increase in weight increases friction.

Thus, the frictional force will change as the crate gets heavier.

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Answer:

Explanation:

First establish the POSITIVE direction.  We'll say to the RIGHT is positive, and UP is positive.  All you have to do is add forces in the positive direction and subtract forces in the negative direction.  If your answer is positive, the net force is to the right or up;  if your answer is negative, the net force is to the left or down.

Here are your answers:

25-15-15 = -5N left

84-60-10 = 14N up

75-60-2 = 13N right

25+25-25 -15 = 10N right

44+44-100= -12N down

67-76 = 9N right

The value of Young's modulus will be 6.78*10^8 N/m^2

Linear elastic solid materials have a mechanical characteristic called Young's modulus (Y), which is also referred to as the elastic modulus. It explains how strain (proportional deformation) and stress (force per unit area) interact with one another in a material.

The stiffness of a solid or its resistance to elastic deformation under stress is measured by the Young's modulus (E or Y). It connects strain (proportional deformation) along an axis or line to stress (force per unit area).

Given elongation delta L=1.10 m

Mass of A=65.0 kg,

L=45.0 m

D=7 mm

radius =  r=7/2=3.5 mm

Since we know the Young's Modulus (Y) Can be defines as [tex]Y=\frac{\text { stress }}{\text { strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta l / L}[/tex]

Now, (4/{L)=(1.10 m/45.0 m)=0.0244

[tex]\begin{aligned}\ \frac{F}{A}=\frac{M g}{A} & =\frac{65 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{sec}^2}{\pi r^2} \\& =\frac{65 \mathrm{~g} \times 9.8 \mathrm{~m} / \mathrm{sec}^2}{3.14 \times\left(3.5 \times 10^{-3}\right)^2} \\& =\frac{65 \times 9.8}{3.14 \times 12.25 \times 10^{-6}} \mathrm{~N} / \mathrm{m}^2 \\& =\frac{637}{38.465} \times 10^6 \mathrm{~N} / \mathrm{m}^2 \\\frac{F}{A} & =16.56 \times 10^6 \mathrm{~N} / \mathrm{m}^2\end{aligned}$$[/tex]

Therefore, the value of Young's modulus will be

[tex]$$\begin{aligned}& Y=\frac{F / A}{\Delta l / L}=\frac{16.56 \times 10^{-6} \mathrm{~N} / \mathrm{m}^2}{0.0244}=\left(678.688 \times 10^6\right) \mathrm{N} / \mathrm{m}^2 \\& Y=6.78 \times 10^8 \mathrm{~N} / \mathrm{m}^2\end{aligned}$$[/tex]

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To throw out a runner, the catcher must toss the ball 130 feet out, 3 feet behind home plate, up to second base.

In order to throw out a runner, the catcher must throw the ball a distance of c=84.85 ft from home base to second base.

The distance between second base and the back tip of home plate shall be 127 feet, 3 inches.

As a result, although though the bases' "points" are 90 feet apart, the actual distance between each subsequent set of base markers is more like 88 feet (26.8 m).

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Answer:

i dont know

Explanation:

The mass of the pilot flying a small plane is 75.74 kg.

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Speed of the small plane: v = 130 m/s.

Radius of circular path: r = 400 meter.

Hence, centripetal acceleration: a = v²/r = (130)²/400  m/s² = 42.25 m/s²

Required force: F = 3200 N.

Hence, from Newton's 2nd law of motion:

The pilot's mass  = force/acceleration

= 3200/42.25 kg

= 75.74 kg.

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Answer:

In parallel combination each appliance gets the full voltage.

If one appliance is switched on/of others are not affected.

The parallel circuit divide the current through the appliances.

In a parallel combination it is very easy to connect or disconnect a new appliance without affecting the working of other appliances.

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Explanation:

Answer:

Centimeters.

Explanation:

There are only 2 viable answers here. This is due to the fact that moles are used not to measure diameters but instead the number of atoms.

Similarly, milligrams are not used to measure diameters but instead small volumes.

This leaves us with kilometers and centimeters, and since all 3 objects are more easily representable in centimeters than kilometers, the answer must be centimeters. One example would be the diameter of a ping pong ball, which is approximately 4cm, or in other words, 0.00004km, with 4cm being much more easy to represent as it overall involves less digits.

Answer:

Option D. It quadruples

Explanation:

Let the initial voltage be V.

Let the initial power be P.

We shall determine the final power of the circuit as follow:

Initial voltage (V₁) = V

Initial power (P₁) = P

Final voltage (V₂) = double the initial voltage = 2P

Final power (P₂) =?

NOTE: Resistor (R) is constant

P = V² / R

Cross multiply

PR = V²

Divide both side by P

R = V² / P

Thus

V²₁ / P₁ = V²₂ / P₂

V² / P = (2V)² / P₂

V² / P = 4V² / P₂

Cross multiply

V² × P₂ = P × 4V²

Divide both side by V²

P₂ = P × 4V² / V²

P₂ = P × 4

P₂ = 4P

Thus, the power will be four times the initial power i.e quadruple.

The force experienced by the ball to get accelerated was found to be 3.65.

The rate of change in velocity is defined as acceleration. The magnitude of acceleration at every point along a trajectory is provided by the rate of change of velocity in both magnitude and direction at that location. The real acceleration at time t is determined in the limit as a time interval.

Time = 20s

The initial velocity is 0m/s.

Final speed = 73m/s

Unknown: Experience with forcing the ball =?

To answer this problem, we use Newton's second law of motion's equation:

formula for force was :

F=ma

The mass is denoted by m.

The final velocity is denoted by v.

The starting velocity is denoted by u.

It is the amount of time spent

So;

F = m (v - u / t)

F = m (73-0/20)

= 3.65 times mass

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The time it takes the sound wave to travel to and fro through 2000 m is 11.66 seconds.

A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium.

To calculate the time it takes the sound wave to travel, we use the formula below

Formula:

Where:

From the question,

Substitute these values into equation 1

The time it takes is 11.66 seconds.

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The first question is asking which phase change occurs when steam is cooled to 100°C. The answer is condensation.

The second question is asking for the term that refers to the amount of energy needed to change a liquid into a gas, without changing its temperature. The answer is heat of vaporization.

The third question is asking in which phase change hydrogen bonds are formed. The answer is freezing.

The fourth question is asking for the amount of energy required to raise the temperature of 20 g of a substance from 72°C to 88°C, given that the specific heat of the substance is 0.215 J/g°C. To solve this problem, you can use the formula Q = m * c * delta T, where Q is the energy required, m is the mass of the substance, c is the specific heat, and delta T is the change in temperature. Plugging in the given values, you get Q = 20 g * 0.215 J/g°C * (88°C - 72°C) = 68.8 J. Therefore, the answer is 68.8 J.

As Projectile range is independent of mass, so They all will have same range.

When an object is close to Earth's surface, its gravitational attraction is calculated using the formula F = mg, where m is the object's mass and g is a constant whose value varies depending on the location. On the surface of the earth, it averages 9.81 meters per second. And when one moves away from the surface in either an inner or an outward direction, this value drops.

Although the strength of the earth's gravitational pull changes depending on the mass of the object, the acceleration caused by it is constant (of course close to the surface). Consequently, when two items of different masses are dropped from the same height, they hit the ground simultaneously (neglect air resistance).

Any motion of a projectile simply consists of vertical and horizontal motion. Regardless of the direction the projectile is going in, gravity controls vertical motion since it only acts in that direction. Additionally, the acceleration caused by gravity is not reliant on the mass of the item. Therefore, the mass has no bearing on the vertical motion.

A projectile moves horizontally in a very straightforward manner. A projectile is not affected by horizontal forces. As a result, its horizontal velocity does not change. Additionally, the range is influenced by horizontal velocity and flying time. Flight time is influenced by vertical motion.

We now know that mass has no bearing on flying duration or range. And if you run into a projectile motion problem, simply divide the problem into horizontal and vertical motions. And handle it the same way you would a single-direction motion. greatly simplifies matters.

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For pendulum A, the period of oscillation is T = 2π × √(L/g). For pendulum B, the period of oscillation is T = 2π × √(2L/g). As we can see from the formula, Pendulum B takes longer for a swing as its period of oscillation is longer than pendulum A due to the fact that half the mass is distributed along the uniform bar.

The period of a pendulum is given by the formula T = 2π × √(L/g) where L is the length of the pendulum and g is the acceleration due to gravity.

For pendulum A, the total mass is concentrated in the small ball at the end of the massless bar, so the period of oscillation is T = 2π × √(L/g)

For pendulum B, half the mass is in the small ball, and half is distributed along the uniform bar, so the period of oscillation will be T = 2π × √(2L/g)

In both cases, the length of the pendulum is the same, so the period of oscillation for pendulum A is shorter than the period of oscillation for pendulum B. Therefore pendulum B takes longer to swing.

It's worth noting that these calculations are based on small oscillations, when the angle of oscillation is small, as the period of a pendulum is dependent on the angle of oscillation.

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According to the question linear speed must an Earth satellite have to be in a circular orbit at an altitude of 204 km above Earth's surface is 30.682× [tex]10^{10}[/tex].

Astronomically speaking, an orbit is the curved path—typically elliptical—that a planet, satellite, comet, etc. takes when it revolves around another celestial body as a result of gravitation.

By Newton's second law

GMm/[tex]r^{2}[/tex] m[tex]v^{2}[/tex] /r.

Since the radius of Earth is 6.37 × [tex]10^{6}[/tex] m,

the orbit radius is r=(6.37×[tex]10^{6}[/tex] m+204×[tex]10^{3}[/tex] m)=1.3×[tex]10^{3}[/tex] m.

The solution for v is

v= [tex]\sqrt{\frac{GM}{r} }[/tex]​ ​ = (6.67×[tex]10^{-11}[/tex]× 5.98×[tex]10^{24}[/tex]/1.3×[tex]10^{3}[/tex])^1/2

⇒ v= 30.682× [tex]10^{10}[/tex]

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Answer:

  A.  35 °C

Explanation:

You want to know which of the temperatures 5, 10, 15, or 35 °C is most favorable to mosquito breeding.

Mosquitos are nonfunctional below 50 °F (10 °C), and are lethargic at 60 °F (15.6 °C). They prefer 29 °C, and their breeding falls off at higher and lower temperatures.

Of the temperature choices offered, mosquitos would breed fastest at 35 °C.

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

The velocity at Point P is 1.72 x 10^(6) m/s.

Mass of proton is Mp = 1.67x10^(-27)

Initial velocity Vo = 2.1x10^6 m/ s

As the proton is moving its Total energy should remain constant throughout the motion to point P.

Conserving the energy,  

Initial energy = Final energy

(1\2)MpVo^2 + (kQ1Q2)/R1 = (1/2)MpV^2 +(kQ1Q2)/R2  

Given R1 = 3.0 mm  

R2 = 4.0 mm  

2kQ1Q2( 1/R1 - 1/ R2) = Mp( V^2 - Vo^2)

2x9x10^9x(-10x10^(-9))x1.6x10^(-19) x {(1000/3)-(1000/4)} = 1.67x10^(-27){ V^2 - Vo^2}  

-2.4x10^(-15) = 1.67x10^(-27){V^2- Vo^2}  

-1.44x10^12 + Vo^2 = V^2  

-1.44x10^12 + (2.1x10^6)^2. = V^2  

2.97x10^(12) = V^2

V = 1.72x10(^6) m/s  

Hence velocity at Point P is 1.72 x 10^(6) m/s.

Proton's speed is dependent on a variety of factors, such as its energy and environment. When protons are released from an accelerator, their speed can be as high as nearly the speed of light, or roughly 300 million meters per second. When protons are found in an environment such as space or the Earth’s atmosphere, their speed is much lower.

In space, protons typically move at around 100 km/s or 0.1% of the speed of light. Within Earth’s atmosphere, protons tend to move much slower, usually around 10 km/s. In the upper atmosphere, protons may be accelerated by solar wind or cosmic rays to speeds of up to 400 km/s. Protons may also be slowed down by collisions with other particles, such as electrons or atoms. In general, the speed of a proton is determined by its energy and the environment it is in.

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Complete Question:

A proton follows the path shown in (Figure 1). Its initial speed is

[tex]V_0[/tex] = 2.1x10^6 m/s.

Part A What is the proton's speed as it passes through point P? Express your answer to two significant figures and include the appropriate units.

[tex]V_P[/tex][tex]= ?[/tex]

Answer:

a. hydrogen sulfide.

Answer:

P V = N R T      is the universal gas equation

If only pressure and volume are considered (the other factors are constant)

P1 V1 = P2 V2

So P2 = (V1 / V2) * P1

This is also known as Boyle's Law:

P V = constant

The mean free path between collisions is 1.41 mm.

The mean free path between collisions in a gas can be calculated using the formula: mean free path = 1 / (density x collision cross-section)

To calculate the mean free path between collisions for the ping-pong balls, we first need to calculate the number density of the balls, which is the number of balls per unit volume. Since the box is 1m x 1m x 1m, and the diameter of a ping-pong ball is 0.03m, the volume of the box is 1m^3 and the volume of a single ball is 4/3 * pi * (0.03/2)^3 = 2.5*10^-5 m^3.

Therefore, the number density of the balls is 2000 / 1 = 2000 balls/m^3.

The collision cross-section of a sphere is given by the formula:

cross-section = pi * (diameter / 2)^2

So, the collision cross-section of a ping-pong ball is pi * (0.03 / 2)^2 = 7.07 * 10^-4 m^2

Therefore, the mean free path between collisions is:

mean free path = 1 / (2000 x 7.07 * 10^-4) = 1.41 * 10^-3 m or 1.41 mm

This means that the ping-pong balls in the lottery machine are colliding on average every 1.41mm.

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A = (v-u)/t, where an is acceleration, v is final speed, u is initial speed, and t is time, is the equation for acceleration. When the values are filled in, we obtain a = (60-20)/5 or an is 8 m/s2.

The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t). This enables you to calculate the change in velocity in meters per second squared (m/s2).

+2 m/s2

If there is no change in velocity between 2 and 3 seconds, the acceleration is 0 m/s2. The acceleration is +2 m/s2 over the first 0 to the second interval. The acceleration is zero between seconds two and three. The acceleration is -3 m/s2 throughout the three to six-second period.

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The work done by the net external force acting on the skier is 3368.56 J.

When an object is moved over a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work in physics.

The joule is the SI unit of labor (J). It is described as the work accomplished over a one-meter distance by a force of one newton.

It is given that,

Mass of the water skier, m = 79.0 kg

Initial speed of the water skier, u = 5.6  m/s

Final speed of the skier, v = 10.8 m/s

Now, we calculate work done by using work energy theorem,

W = ΔE

[tex]W=\frac{1}{2}m (v^2-u^2)[/tex]

[tex]W=\frac{1}{2} 79.0 kg (10.8^2-5.6^2)[/tex]

[tex]W=3368.56J[/tex]

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Answer:

10 kg

Explanation:

Use law of conservation of momentum

momentum of system before collision = momentum of system after collision

    5 kg * 20 m/s +  x kg  * 10 m/s         = 5 kg * 10 m/s   +  x kg * 15 m/s

           100  + 10x                                   = 50 + 15 x

                 50                                =    5x

                   x = 10 kg

  The correct definition of minerals is 'nutrients that strengthen bones and teeth, help keep the blood healthy, and keep the heart and other organs working properly'

 Minerals and vitamins are essential micronutrients that the body uses to carry out its normal functions. Minerals can be obtained from foods (or artificially), these micronutrients have fundamental roles in growth, fluid balance, bone health, and different physiological processes.  Some of the most important minerals include calcium, iron, phosphorous, fluoride, selenium, sodium, copper, iodine, and zinc.