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soldier1979 [14.2K]

4 years ago

9

Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface?

2 answers:

marshall27 [118]4 years ago

4 0

:Sample Response: Jupiter would have the strongest gravitational pull on a satellite orbiting above its surface because gravity is directly proportional to mass and Jupiter is the most massive planet.

Explanation:

8090 [49]4 years ago

3 0

According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.

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Planets that are cold have only a small amount of gravity

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In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

$u=+6.3 m/s$ is the initial velocity (positive because it is upward)

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

$v=+6.3 + (-9.8)(1.7)=-10.4 m/s$

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

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3 years ago

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The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

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3 years ago

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