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erica [24]
3 years ago
12

A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If

air resistance does –1.8 J of work on the ball, how much energy goes into heating the ground and the ball when it bounces?
Physics
1 answer:
Gala2k [10]3 years ago
5 0

Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

Let g = 9.81 m/s2

\Delta E_p = mgH - mgh = mg(H - h) = 0.32*9.81*(2 - 1.2) = 2.51 J

Since 1.8J of 2.51 J is due to work by air resistance, the rest of the energy (2.51 - 1.8 = 0.71 J) is would go to heating in the ground and ball when it bounces.

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10. CALCULATE: Find the amount of power used in each of the following examples. Show your calculations. a. You use a force of 10
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Answer:

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Explanation:

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3 0
3 years ago
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
3 years ago
A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi
WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

m1v1 + m2v2 = m1'v1' + m2'v2'

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

0 + 0 = 91*v1' + 1300*0.17

v1' = -1300*0.17/91 = -2.4286\ m/s

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

distance = 2.4286 * 7.5 = 18.2145\ meters

8 0
3 years ago
0.0884 moles of a diatomic gas
Sloan [31]

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

\Delta\ U = nC_{v}\Delta\ T

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J

Now, we can apply First Law of Thermodynamics as follows:

\Delta\ Q = \Delta\ U + W

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>

7 0
3 years ago
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