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4 years ago

If a 2 kg ball is traveling at a speed of 4 m/s, what is its kinetic energy?

Physics

2 answers:

Kay [80]4 years ago

8 0

Answer: 16J

Explanation:

K.E = $\frac{1}{2}$ m $v^{2}$

K.E = $\frac{1}{2}$ X 2 X 16

K.E = 16

Art [367]4 years ago

4 0

The kinetic energy of the ball is 16 J.

Answer: Option D

<u>Explanation: </u>

Kinetic energy is the form of energy exhibited by a body when it is in motion. The kinetic energy will be directly proportional to the mass of the body as well as to the square of the speed of the body at which it is moving. So the mathematical representation of kinetic energy of any object is

$\text {Kinetic energy}=\frac{1}{2} \times m \times(v)^{2}$

In the present case, the ball is the object with mass of around 2 kg and it is moving with the speed of about 4 m/s. So, the kinetic energy exhibited by the ball during its motion is

$\text {Kinetic energy of the ball}=\frac{1}{2} \times 2 \times(4)^{2}=16 \mathrm{J}$

So, the kinetic energy exhibited by the ball is 16 J.

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Im pretty sure it’s a because it makes more sense you know?.

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3 years ago

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yuradex [85]

1,) C

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4 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

8 0

2 years ago

A car initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a c

Kitty [74]

We calculate the difference between the two velocities,
135 km/h - 35 km/h = 100 km/h
Then, convert the difference into units of m/s
100 km/h x (1000 m/ 1 km) x (1 hour / 3600 s) = 27.78 m/s
Then divide the difference by time.
acceleration = 27.78 m/s / 10.5 s = 2.65 m/s²

(2) In terms of g,
x = 2.65 / 9.80 = 0.27
Therefore, the answer would be,
a = 0.27g

6 0

4 years ago

Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?

Rzqust [24]

If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer is the option D. only at constant accelerations.

3 0

4 years ago