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2 years ago

6

a car decelerates at a rate of 2.0 m/s^2 and comes to a complete stop after traveling 25 m. What was the speed of the car right

before declaration?

1 answer:

cricket20 [7]2 years ago

8 0

Answer:

${v}^{2} = {u}^{2} + 2as \\ {0}^{2} = {u}^{2} + 2 (- 2)(25) \\ {u}^{2} = 100 \\ \therefore \: u = 10 \: m .{s}^{ - 1}$

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Please find the explanation below

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Which of the following types of radiation can penetrate through paper but not through wood?

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2 years ago

A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

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