2 years ago

A cameraman sitting near the open door of a news helicopter accidentally drops his 140-g mobile phone out

Physics

2 answers:

Debora [2.8K]2 years ago

6 0

Answer:

140m/s

Explanation:

Given parameters:

Mass of mobile phone = 140g

Altitude = 990m

Unknown:

Final velocity = ?

Solution:

This is a motion problem. It is solvable by using the right motion equation. For this problem, understand that the phone is dropping under the influence of gravity.

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height of fall

Initial velocity is 0

Input the variables;

V² = 0² + 2(9.8)(990)

V = 139.29m/s = 140m/s

Agata [3.3K]2 years ago

4 0

Answer:B.140 m/s

Explanation:

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sergey [27]

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2 years ago

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

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Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$