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3 years ago

10

What is the force of a lead ball that starts with a momentum of 7 kgm/s and end with a momentum of 22 kgm/s over a time of 31

seconds?
Explain how you got your answer.

1 answer:

Gwar [14]3 years ago

6 0

Answer:

F = 0.483 N

Explanation:

Initial momentum, $P_i= 7\ kg-m/s$

Final momentum, $P_f= 22\ kg-m/s$

Time, t = 31 s

We need to find the force of a lead ball. We can use here the impulse momentum theorem.

$\text{Change in momentum}=F\times t$

F is force

$P_f-P_i=Ft\\\\F=\dfrac{P_f-P_i}{t}\\\\F=\dfrac{22-7}{31}\\\\F=0.483\ N$

So, the force is 0.483 N.

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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul

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The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$ = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}\\\\$

$= \frac{126* 2.70}{1800}$

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Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

= 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

= 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

$= \frac{1}{2}$ × 126 × $(2.70) ^2$

= 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

= $\frac{1}{2}$ ×1800× $(0.189) ^2$

= 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

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Answer:

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(b)Calculate the magnitude of the acceleration of each object when released.

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Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

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Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

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$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

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