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2 years ago

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Physics

2 answers:

loris [4]2 years ago

7 0

Answer:

$v_f = 17.15 m/s$

Explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

$u = 0$

acceleration of the ball due to gravity is given as

$a = 9.8 m/s^2$

now the final speed of the ball is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.8)(15)$

$v_f = 17.15 m/s$

Sergeeva-Olga [200]2 years ago

4 0

By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s

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Answer:

option B

Explanation:

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$\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

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$F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

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hence, the correct answer is option B

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2 years ago

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

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Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

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The potential energy gained by the rock at the top of the cliff = mgh.

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