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kobusy [5.1K]
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
U
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:
MaRussiya [10]2 years ago
3 0

Explanation:

(10) Mass of a soccer player, m = 0.42 kg

Initial speed, u = 0

Final speed, v = 32.5 m/s

Time, t = 0.21 s

We need to find the force that sends soccer ball towards the goal.

Force, F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N

So, 65 N of force soccer ball sends towards the goal.

(11) Mass of the satellite, m = 72,000 kg

Initial speed, u = 0 m/s

Final speed, v = 0.63 m/s

Time, t = 1296 s

We need to find the force is exerted by the rocket on the satellite.

Force, F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N

So, 35 N of the force is exerted by the rocket on the satellite.

Hence, this is the required solution.

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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
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(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

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U_i is the gravitational potential energy at the surface

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So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

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