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Ksenya-84 [330]
4 years ago
12

A machine performs 482 J of work over a period of 1.5 min. How much power does the machine have?

Physics
1 answer:
stira [4]4 years ago
7 0
Power = Energy / time

P = 482J/90s = 5.3556W

P = 5.3556W
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If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite directi
Lelu [443]

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

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Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

<u>ΔP = 14.5 Ns</u>

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

<u>I = 14.5 Ns</u>

the average force can be found as:

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ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>

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3 years ago
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

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distance traveled d=77.3 cm

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mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

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x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

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