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3 years ago

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A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a velocity of 8 m/s. What is the velocity of t

he skater immediately after?

1 answer:

KengaRu [80]3 years ago

5 0

Answer:

The velocity will be v1 = 0.58[m/s]

Explanation:

This problem can be solved by the law of conservation of the moment, which explains that the moment of a system remains constant because there are no external forces acting on it.

We have the following initial data:

m1 = mass of the skater = 55 [kg]

m2 = mass of the ball = 3 [kg]

v2 = velocity of the ball = 8 [m/s]

Therefore:

$m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}+m_{2}*v_{2}\\(50*0)+(3*0)=(50*v_{1})+(3*8)\\50+3-24=50*v_{1}\\v_{1}= 0.58[m/s]$

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Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

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Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

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3 years ago

A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .

xenn [34]

Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination $\theta=8.0^{\circ}$

Coefficient of friction factor $\mu k=0.056$

To find:

How long it takes to reach the bottom ‘t’

<u>Step by Step Explanation:</u>

Solution:

We know that the formula for weight of the soap bar is given as

$F_{g}=m g \sin \theta$

The frictional force acting on this soap bar is determined by

$F_{f}=\mu m g \cos \theta$

To determine the constant acceleration of the bar, we derive as

$F=m a$

Here $F=F_{g}-F_{f}$ and thus

$F_{g}-F_{f}=m a$ $m g \sin \theta-\mu m g \cos \theta=m a$

$a=g \sin \theta-\mu g \cos \theta$

Where $F_{g}$ =Force imparted due to weight

$F_{f}$ =Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

$\sin \theta$ and $\cos \theta$ are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

$s=\frac{1}{2} a t^{2}$ , From this

$a t^{2}=2 s$

$t^{2}=\frac{2 s}{a}$

$t=\sqrt{\frac{2 s}{a}}$

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

$t=\sqrt{\frac{2 \times 9.2}{a}}$ and we know that

$a=g \sin \theta-\mu g \cos \theta$

$=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0$

$=1.364-0.543$

$a=0.821$

$t=\sqrt{\frac{2 \times 9.2}{0.821}}$

$t=\sqrt{\frac{19.6}{0.821}}$

$t=\sqrt{23.87332}$

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

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3 years ago

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