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2 years ago

14

A freight train is accelerating on a level track. The tension in the coupling between the engine and the first freight car would

change if some of the cargo in the last car were transfered to other cars?

1 answer:

rewona [7]2 years ago

6 0

Answer:

False

Explanation:

According to Newton's 3rd law, the tension between the engine and the 1st freight cars would equals to the force required to pull all the freight cars from the 1st one on ward. In addition, Newton's 2nd law states that the F = ma, which means that force is proportional to the mass of all the freight cars, given that the freight train is accelerating. So if some of the cargo in the last car were transferred to other cars, their total mass would stay the same, and so as force F and tension.

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A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

Tasya [4]

<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>

Work is the product of force and the distance covered. Therefore, Work = force × distance.

Work is measured in joules.
Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

= 1/2 × 10× 5²

= 5 × 25

= 125 joules

Hence, work done is 125 joules.

3 0

2 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago

In which of the following examples does the object have both kinetic and potential energy? Select all that apply.

Andru [333]

Objects having both kinetic and potential energy:

water flowing downstream

a child swinging on a swing

a bouncing ball

a plane in flight at 30,000 feet

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. It is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Therefore, an object has kinetic energy when its speed is non-zero (so, whenever it is moving).

The potential energy of an object is the energy possessed by the object due to its position in the gravitational field. It is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the heigth of the object relative to the ground

Therefore, an object has potential energy whenever it is located at a certain height above the ground.

So in this problem, the objects that have both kinetic and potential energy are:

a rock at the edge of a cliff --> NO, because the rock is at rest (so KE = 0)

water flowing downstream --> YES, because the water is moving AND it is at a certain height above the ground

a child swinging on a swing --> YES, because the child is moving AND it is at a certain height above the ground

a bouncing ball --> YES, because the ball is moving AND it is at a certain height above the ground

water behind a dam --> NO, because the water is at rest (so KE=0)

a car moving on a level road --> NO, because the car is at ground level (so PE=0)

a plane in flight at 30,000 feet --> YES, because the plane is moving AND it is at a certain height above the ground

a compressed spring --> NO, because the spring is at rest (so KE=0)

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0

2 years ago

In an experiment to illustrate the propagation of sound through fluids, identical sound sources emitting at 440 Hz are used to p

earnstyle [38]

Answer:

B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.

Explanation:

To solve this easily, we can just calculate the wavelength of the sound in Ethanol and in Mercury.

In Ethanol, the wavelength will be:

λ = c/f

λ = 1160/440

λ = 2.63 m

In Mercury, the wavelength will be:

λ = c/f

λ = 1450/440

λ = 3.3 m

The wavelength of sound is greater in Mercury than in Ethanol but the frequency is the same.

Frequency of sound is not dependent on medium, but velocity and wavelength change depending on the medium.

3 0

2 years ago