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3 years ago

A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground

Physics

2 answers:

pogonyaev3 years ago

7 0

T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s

Paraphin [41]3 years ago

5 0

V = u + at
u=0(rest) ; a=gravitational acc(g)=10
v = 0 + 10(3.5) = 35m/s

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3 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

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3 years ago

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

jeka94

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

$m_av_a + m_cv_c = 0$

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

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

$= - 0.189$ m/s

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.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

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2 years ago