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3 years ago

6

A student has 67-cm-long arms. What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle

without spilling any? The distance from the handle to the bottom of the bucket is 35 cm .

2 answers:

choli [55]3 years ago

8 0

Answer:

29.61 rpm.

Explanation:

Given,

student arm length, l = 67 cm

distance of the bucket, r = 35 m

Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.

Now,

$mg = m r \omega^2$

$\omega = \sqrt{\dfrac{g}{R}}$

R = 67 + 35 = 102 cm = 1.02 m

$\omega = \sqrt{\dfrac{9.81}{1.02}}$

$\omega = 3.101\ rad/s$

$\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s$

$\omega = 0.494 \times 60 = 29.61\ rpm$

minimum angular velocity is equal to 29.61 rpm.

Damm [24]3 years ago

5 0

Answer:

29.6 rpm

Explanation:

length of arm = 67 cm

distance of handle to the bottom = 35 cm

radius of rotation, R = 67 + 35 = 102 cm = 1.02 m

The centripetal force acting on the bucket is balanced by the weight of the bucket.

mRω² = mg

R x ω² = g

$\omega = \sqrt\frac{g}{R}$

$\omega = \sqrt\frac{9.8}{1.02}$

ω = 3.1 rad/s

Let f is the frequency in rps

ω = 2 x 3.14 x f

3.1 = 2 x 3.14 xf

f = 0.495 rps

f = 29.6 rpm

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2 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

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3 years ago

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Answer:

Explanation:

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Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

P.E=mgh

P.E=0.0961×9.81×0.6385403

P.E=0.602J

From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another

Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

k=532.72N/m

The spring constant is 532.72 N/m

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3 years ago

How many seconds will it take to travel 3,600 meters if your speed is 90 meters per second?

klio [65]

40

Because if you divide 3,600 by 90 it is equivalent to 40.

Hope this helps, have a blessed day :)

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3 years ago