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olchik [2.2K]
3 years ago
9

Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc

ounters no air resistance?
Physics
2 answers:
tamaranim1 [39]3 years ago
5 0

Answer:

50 m/s

Explanation:

u = 0 m/s, t = 5 seconds, v = ? , g = 10 m/s^2

Use first equation of motion

v = u + g t

v = 0 + 10 x 5 = 50 m/s

Tasya [4]3 years ago
3 0

Answer:

The rock's speed after 5 seconds is 98 m/s.

Explanation:

A rock is dropped off a cliff.

It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².

Speed after 5 seconds = V

We know that acceleration = average speed/time

In our case,

g = ((0+V)/2)/5

9.8*5 = V/2

=> V = 2*9.8*5

V = 98 m/s

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Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

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Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

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Explanation:

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Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the
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Answer:

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Explanation:

Part a)

As we know that there is no external torque on the system of two twins

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Part b)

Since angular momentum is conserved here as there is no external torque

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