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KiRa [710]
3 years ago
12

A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears t

he top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.
Physics
1 answer:
yarga [219]3 years ago
8 0

Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s

The average velocity is 197.263157895 m/s

v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s

The average velocity is 169.491525424 m/s

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Explanation:

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We can see that the more the mass of a substance, the higher the specific heat needed to raise the temperature.

Specific heat is directly proportional to the mass of substance.

It also dependent on the amount of heat supplied and the nature of the substance.

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Also, the capacitance of a capacitor c = q/v where

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