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hodyreva [135]
2 years ago
15

10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends

soccer ball towards the goal.
G
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E
S
S
Formula

11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?

G
U
E
S
S
Formula

please I need help I don't understand it and I had to deliver it yesterday helpp:(

Physics
1 answer:
maxonik [38]2 years ago
5 0

Answer:

10. 65 N

11. 35 N

Explanation:

10. Determination the force that sends the soccer ball towards the goal.

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Initial velocity (u) of the ball = 0 m/s

Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s

Acceleration (a) of the ball =?

a = (v – u) /t

a = (32.5 – 0) / 0.21

a = 32.5 / 0.21

a = 154.76 m/s²

Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:

Mass (m) of the ball = 0.42 kg

Acceleration (a) of the ball = 154.76 m/s²

Force (F) =?

F = ma

F = 0.42 × 154.76

F = 65 N

Thus, the force that sends soccer ball towards the goal is 65 N

11. Determination of the force exerted by the rocket on the satellite.

We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:

Initial velocity (u) of satellite = 0 m/s

Final velocity (v) of satellite = 0.63 m/s

Time (t) = 1296 s

Acceleration (a) of the satellite =?

a = (v – u) /t

a = (0.63 – 0) / 1296

a = 0.63 / 1296

a = 4.861×10¯⁴ m/s²

Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:

Mass (m) of the satellite = 72000 Kg

Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²

Force (F) =?

F = ma

F = 72000 × 4.861×10¯⁴

F = 35 N

Thus, the force exerted by the rocket on the satellite is 35 N

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Explanation:

Given that :

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the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

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For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = C_v=\dfrac{3}{2}R

For Part A:

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K

Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)

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Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K

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Q=218.2425 \ J

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A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time =  20 μs = 20 * 10^{-6} s

let distance to the reflecting surface be = x

total distance travelled by pulse will be  = 2x

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distance = speed * time

2x = 3.0 × 10^{8} * 20 * 10^{-6}

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