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alexgriva [62]
3 years ago
5

A student drops a 2.4-kg ball. It’s speed right before hitting the ground is 5 m/s. If 15 J of energy was transferred to the env

ironment as the ball fell, what was the total initial energy of the ball?
21 J
30 J
45 J
6 J
Physics
1 answer:
Travka [436]3 years ago
8 0

The answer is 45J

Don’t know how to explain sorry

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The separation between two protons is increased by a factor of 2. How does the force between them change?
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Decrease by a factor of 2
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when a stationary rugby ball is kicked, it is contact with a player's about for 0.05 s. during this short time, the ball acceler
Ad libitum [116K]

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

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Which of the following is produced by moving parts of machine?
AnnZ [28]
The answer is Electricity/heat
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Calculate the volume that 42g of nitrogen gas (N2)occupies at standard temperature and pressure.​
luda_lava [24]

Nitrogen gas has a molar mass of about 28.0134 g/mol. Then we have a starting amount of

(42 g) / (28.0134 g/mol) ≈ 1.4993 mol

of N₂.

At standard temperature and pressure, one mole of an ideal gas occupies a volume of about 22.4 L. Then 42 g, or 1.4993 mol, of N₂ takes up

(1.4993 mol) × (22.4 L/mol) ≈ 33.6 L

8 0
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A billiard ball with a mass of 1.5kg is moving at 25m/s and strikes a second ball with a mass of 2.3 kg that is motionless. Find
aliya0001 [1]

Answer:

The velocity of the second ball after the collision will be 16.3 m/s.

Step-by-Step Explanation:

Law of conservation of momentum states that the total momentum of an isolated remains the same before and after the collision.

Let: mass of the first ball = m1 = 1.5 kg

mass of the second ball = m2 =  2.3 kg

Momentum before the collision:

velocity of the first ball = v1 = 25 m/s

velocity of the second ball = v2 = 0 m/s

[tex] Initial momentum = m1v1 + m2v2 = 1.5*25 + 2.3*0 = 37.5 kgm/s [\tex]

Momentum after the collision:

velocity of the first ball = v1 = 0 m/s

velocity of the second ball = v2 = x

[tex]momentum after the collision= m1v1 + m2v2 = 1.5*0 + 2.3*x = 2.3*x[\tex]

According to law of conservation of momentum:

Initial momentum = Momentum after collision

37.5 = 2.3*x

⇒ x = 37.5/2.3

x = 16.3 m/s

4 0
3 years ago
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