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AysviL [449]
3 years ago
13

If I drop a paper clip from rest and it falls freely until it hits the ground with a speed of 30 m/s. How long is the ball in fr

ee fall?​
Physics
1 answer:
enot [183]3 years ago
4 0

Answer:

<h3>3.06s</h3>

Explanation:

If I drop a paper clip from rest and it falls freely until it hits the ground with a speed of 30 m/s,we ae to find the time taken for the ball to drop. To do that we will use the equation of motion v = u + gt

v is the final velocity = 30m/s

u is the initial velocity = 0m/s (the object drops from rest)

g is the acceleration due to gravity = 9.81m/s²

time is the time taken

Substituting the given parameters into the formula;

30 = 0 + (9.81)t

30 = 9.81t

t = 30/9.81

t = 3.06secs

<em></em>

<em>Hence it will take the object 3.06secs in free fall</em>

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A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of
garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
3 years ago
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