All categories

3 years ago

13. A lever does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's

Physics

1 answer:

shusha [124]3 years ago

6 0

Answer:

The resulting speed of the ball is;

10 m/s

Explanation:

The given parameters are;

The amount of work done by the lever = 5.0 J

The mass of the ball on which the work is done, m = 0.10 kg

By the principle of conservation of energy, we have;

The work done by the lever = The kinetic energy. K.E., gained by the ball

The kinetic energy, K.E., is given by K.E. = 1/2·m·v²

Where, for the ball, we have;

m = The mass of the ball

v = The resulting speed of the ball

Therefore, by substituting the known values, we have;

The work done by the lever = 5.0 J = The kinetic energy. K.E., gained by the ball = 1/2 × m × v² = 1/2 × 0.10 kg × v²

∴ 5.0 J = 1/2 × 0.10 kg × v²

v² = 5.0 J/(1/2 × 0.10 kg) = 100 m²/s²

v = √(100 m²/s²) = 10 m/s

The resulting speed of the ball, v = 10 m/s.

You might be interested in

What do the arrows at point three indicate

Maru [420]

Maybe show a picture ? I don’t get the question .

5 0

3 years ago

If the speed of an object is increasing, then the forces acting on that object must be ________.

In-s [12.5K]

Answer:

Unbalanced.

Explanation:

Usually, unbalanced forces cause acceleration, or increased movement.

5 0

3 years ago

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a thi

Drupady [299]

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following

Explanation:

After the full moon (maximum illumination), the light continually decreases. So the waning gibbous phase occurs next. Following the third quarter is the waning crescent, which wanes until the light is completely gone -- a new moon.

waning gibbous phase

The waning gibbous phase occurs between the full moon and third quarter phases. The last quarter moon (or a half moon) is when half of the lit portion of the Moon is visible after the waning gibbous phase.

Time takes by the moon to go through all the phases

about 29.5 days

It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).

At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!

6 0

3 years ago

Read 2 more answers

A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

cestrela7 [59]

Hope this helps you!

7 0

3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

Anna71 [15]

Answer:

B. $V_{f}= 10\,cubic\,inches$

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

$PV=nRT$

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

$P_{i}V_{i}=n_{i}RT_{i}$ (1)

$P_{f}V_{f}=n_{f}RT_{f}$ (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because $T_{i}=T_{f}$ , $n_{i}=n_{f}$ and R is an universal constant:

$P_{i}V_{i}= P_{f}V_{f}$ , solving for $V_{f}$

$V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}$

$V_{f}= 10 cubic\,inches$

6 0

3 years ago