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UNO [17]
3 years ago
14

The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560

g. Find the average power delivered to the train during its acceleration.
Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

= 5.1 W

Explanation:

time (t) = 30 ms = 0.03 s

mass (m) = 560 g = 0.56 kg

initial velocity (U) = 0 m/s

final velocity (V) = 0.74 m/s

power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v

m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}

m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}

= 5.1 W

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6/2times500=1500

Explanation:

s = d/t which means speed equals distance divided by time.

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If Manuel exerts a force of 10 N to push a desk to the right at the same time Lynn exerts a force of 15 N to push the desk to th
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Answer:

The desk will move to the left.

Explanation:

Since Manuel is exerting a force of 10 N to the right, and Lynn is exerting a force of 15 N to the left, it means that Lynn is exerting a force of 5 N more than Manuel, meaning that the desk will move to the left, the direction Lynn is pushing it too.

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2 years ago
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2 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen
bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

f_{a} =(\frac{v+v_{c} }{v} )f_{s} (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s} (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s

b) Substitute eq. 1 in eq. 2:

f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz

7 0
3 years ago
You have a meteorite sample and you decide to use the uranium-235/lead-207 system to date it. After analysis, you find that it h
storchak [24]

Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

% = 2.25 / 150 = 1.5 %      of 235 U left

5 0
2 years ago
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