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4 years ago

Which of the following statements is false?

Physics

1 answer:

astraxan [27]4 years ago

5 0

Answer:

c. Kinetic energy is conserved in all collisions .

Explanation:

the total kinetic energy of system of particles involved in the collision does changes .

kinetic energy is not conserved in all collisions because in some cases energy is converted to another form of energy that is heat etc .

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From the top of a cliff overlooking a lake, a person throws two stones. The two stones have identical initial speeds of v0 = 13.

Alina [70]

Answer:

X = 15.88 m

Explanation:

Given:

Initial Velocity V₀ = 13.4 m/s

θ = 30.1 °

g = 9.8 m/s²

To Find horizontal distance let "X" we have to time t first.

so from motion 2nd equation at Height h = 0

h = V₀y t + 1/2 (-g) t ² (ay = -g)

0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t² (V₀y = V₀ Sin θ)

⇒ t = 1.37 s

Now For Horizontal distance X, ax =0m/s²

X = V₀x t + 1/2 (ax) t ²

X = 13.4 m × cos 30.1° x 1.37 s + 0

X = 15.88 m

8 0

4 years ago

Sever21 [200]

Three units deformations observed in these bands.

<h3>What forces do a rubber band encounter?</h3>

Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.

The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.

Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.

learn more about Elastic force refer

brainly.com/question/5055063

#SPJ14

7 0

1 year ago

A sinusoidal wave travels along a string. if the time for a particular point to move from maximum displacement to zero displacem

Lisa [10]

The time it takes for a point to reach 0 displacement from maximum displacement is $\frac{\pi}{2}$ of a cycle. Thus, one period ( $2\pi$ ) will take $T=0.17*4=0.68s$ .

The frequency is simply $f=\frac{1}{T}=\frac{1}{0.68}=1.47Hz$ .

The speed is given by: $v=f\lambda=1.47*1.4 = 2.06m/s$

7 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

At a given instant in time, a 6-kg rock that has been dropped from a high cliff experiences a force of air resistance of 15 n. w

Vinil7 [7]

Gravitational force is equal to 6 kg * 9.81 m/s^2 = 58.86 N. If air resistance is 15 n up, the magniture is 43.86 N downward.

5 0

3 years ago