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Vlad [161]
2 years ago
7

A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

of mass 500g is thrown upwards with a velocity of 15m/s. Calculate its momentum at the highest point.

Physics
1 answer:
shepuryov [24]2 years ago
5 0

Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>

<em>F</em><em>=</em><em> </em><em>m.a</em>

<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>

<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>

<em>a</em><em>=</em><em> </em><em>acceleration</em><em> </em><em>.</em>

<em>And</em><em> </em><em>the</em><em> </em><em>solutions</em><em> </em><em>are</em><em> </em><em>in</em><em> </em><em>pictures</em><em>. </em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
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Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

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Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



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If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.
elena-14-01-66 [18.8K]
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