What is kinetic energy? (99 points)

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table. A

Annette [7]

The magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Impulse:

The effect of applied force acting for a very small interval of time is known as Impulse.

Given data:

The initial magnitude of the force on Ball A is, F = 50 N.

The time interval for the applied force on Ball A is, t = 1.25 s.

And the impulse at initial on ball A is, I = 62.5 N-s.

Now, if force on ball A doubles, which means new magnitude of force is,

F' = 2F

F' = 2 × 50

F' = 100 N

For same time interval, the new impulse on Ball A is given as,

I' = F' × t

Solving as,

I' = 100 × 1.25

I' = 125 N-s

Taking the ratio of both the impulses as,

I' / I = 125/62.5

I' = 2 I

Thus, we can conclude that the magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Learn more about the impulse here:

brainly.com/question/904448

7 0

3 years ago

A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a

anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = $4.83 \AA$

Intercepts along the three axes:

$\bar{x} = 9.66 \AA$

$\bar{x} = 19.32 \AA$

$\bar{x} = 14.49 \AA$

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = $\frac{1}{9.66}$

b = $\frac{1}{19.32}$

c = $\frac{1}{14.49}$

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = $\frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}$

b' = $\frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}$

c' = $\frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}$

7 0

3 years ago

I really need the answer to this question please

Sliva [168]

I believe the answer is option A

5 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

K 12!! When an observer is moving towards a stationary object making a sound, the pitch of the sound is perceived to be ________

lozanna [386]

When an observer is moving towards a stationary object making a sound,
he runs the risk of startling the stationary object.

On the other hand, if the stationary object is the one making the sound,
then the pitch of the sound perceived by the observer is higher than the
sound that the object is actually making.

4 0

4 years ago