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pashok25 [27]
3 years ago
6

Two objects, one of mass m and the other of mass 2m, are dropped from the top of a building. If there is no air resistance, when

they hit the ground A) both will have the same kinetic energy. B) the heavier one will have twice the kinetic energy of the lighter one. C) the heavier one will have four times the kinetic energy of the lighter one. D) the heavier one will have half the kinetic energy of the lighter one. E) the heavier one will have one-fourth the kinetic energy of the lighter one.
Physics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

option b

Explanation:

the heavier one will have twice the kinetic energy of the lighter one

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A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to
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Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius r=\frac{d}{2}=\frac{15.3}{2}=7.65cm

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Time for change in magnetic field dt = 0.26 sec

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3 years ago
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
ziro4ka [17]

Answer:

25 m/s

Explanation:

from the question you van see that some detail is missing, however i found this same question using internet search engines on: 'https://www.chegg.com/homework-help/questions-and-answers/light-rail-passenger-trains-provide-transportation-within-cities-speed-slow-nearly-constan-q5808369'

here is the complete question:

'Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0m/s . Once free of this area, it speeds up to 12m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final Speed?'

SOLUTION

initial speed (u) = 7 m/s

final speed (v) = 13 m/s

initial acceleration time (t1) = 8 s

final acceleration time (t2) = 16 s

what is the higher cruising speed?

acceleration = \frac{final speed (v) - initial speed(u)}{time (t1)}

acceleration = \frac{13-7}{8} = 0.75 m/s^{2}

since the train accelerates at the same rate, the increase in  speed will be = acceleration x time (t2)

= 0.75 x 16 = 12 m/s

therefore the higher cruising speed = increase in speed + initial speed

= 12 + 13 = 25 m/s

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