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Tema [17]
3 years ago
14

A car moving with a velocity 15m/s and accelerating by 5m/s² attempts to reach a car moving with 30 m/s velocity.What distance s

hould it cover to reach the velocity of other car?
A)30 m
B) 90 m
C) 67 m
D) 70m​
Physics
1 answer:
Firlakuza [10]3 years ago
4 0
I think A is correct
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3 years ago
You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the
AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:              

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

  W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F=  mgsinθ

The work done by you on the suitcase is

W = F L cos0°  ,    where L is he length of the staircase.

W = FL = mgsinθL ,  by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P =\frac{W}{t} =\frac{FL}{t}

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ  eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ    eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

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Answer:

The answer is B.

Explanation:

I meant B. not C so sorry

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