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VikaD [51]
2 years ago
11

A 40.0 kg physics student at rest on a frictionless rink throws a 3.0 kg box, giving the box a velocity of 8.0 m/s. Which statem

ent describes the initial velocity of the student after the box is thrown?
a. 0.6 m/s in the same direction as the box.
b. 0.6 m/s in the opposite direction of the box.
c. 1.7 m/s in the same direction as the box.
d. 1.7 m/s in the opposite direction of the box
Physics
1 answer:
zvonat [6]2 years ago
6 0

Answer: (b)

Explanation:

Given

mass of Student M=40 kg

mass of box m=3 kg

velocity of box v=8 m/s

suppose u is the velocity of the student after the throw

As there is no external force is applied so momentum is conserved

Initial momentum=final momentum=0

0=Mu+mv

Mu=-mv

u=-\frac{3}{40}\times 8=-\frac{3}{5}\\u=-0.6\ m/s

Here negative sign indicates the velocity is opposite in direction i.e. opposite to the direction of the box

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Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your c
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Answer:

d=9.462×10^15 meters

Explanation:

<u>Relation between distance, temps and velocity:</u>

d=v*t

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So:

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Answer:

v=2.24\times 10^{8}\ m/s

Explanation:

Given that,

The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is given by :

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Where

k is the dielectric constant of the substance.

v is the speed of light in water

c=\dfrac{1}{\sqrt{1.78\times 4\pi \times 10^{-7}\times 8.85\times 10^{-12}}}

v=2.24\times 10^{8}\ m/s

So, the speed of light in water is 2.24\times 10^{8}\ m/s

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"Frequency" just means "often-ness" ... how often something happens.
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