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Vladimir [108]
3 years ago
14

You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the

ball reaches relative to the throwing point How long does it take the ball to reach the height What is the position of the ball at t=2s? At what height does the ball have a speed of +5m/s?
Physics
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

Explanation:

Given

initial velocity u=11.2\ m/s

At maximum height velocity of ball is zero

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)^2 -(11.2)^2=2\times (-9.8)\times (s)

s=\frac{11.2^2}{2\times 9.8}

s=6.4

time taken by the ball to reach the maximum height

v=u+at

0=11.2-9.8\times t

t=\frac{11.2}{9.8}

t=1.142\ s

At t=2\ s height of ball is

h=ut+\\frac{1}{2}at^2

h=11.2\times 2-\frac{1}{2}9.8\times (2)^2

h=22.4-19.6

h=2.8\ m

i.e. ball is moving downward

height at v=5\ m/s

v^2-u^2=2as

s=\frac{25-125.4}{2\times (-9.8)}

s=5.12\ m    

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Which choice shows the correct sequence of features formed by continued wave erosion?
attashe74 [19]

Answer:

Wave-cut cliff, sea arch, sea stacks

Explanation:

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The high areas of land adjacent to the incoming wave develop the early features or the formation of wave action, which includes the <em>wave-cut cliff</em>

The continuous undercutting of the cliff by the wave results in the formation of the <em>wave cut platform</em>

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What is exothermic .Is it a reaction?
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An auto race takes place on a circular track. A car completes one lap in a time of 18.0 s, with an average tangential speed of 4
eduard

Answer:

(A) 0.3488 rad/sec

(B) 124.246 m                

Explanation:

We have given car completes one revolution in 18 sec'

So time period T= 18 sec

Tangential speed v = 43.4 m/sec

(A) Angular velocity is given by \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{18}=0.3488rad/sec

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We know that v=\omega r

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A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s​2 until its velocity reaches 500 m/s.
katrin2010 [14]

Answer:

The distance the plane covered while it was accelerating is 80,633.3 m

Explanation:

Given;

initial velocity of the plane, u = 90 m/s

acceleration of the plane, a = 1.5 m/s²

final velocity of the plane, v = 500 m/s

The distance covered by the plane is given as;

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where;

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d = 241900 / 3

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Therefore, the distance the plane covered while it was accelerating is 80,633.3 m

5 0
3 years ago
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