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4 years ago

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The uniform disk of mass m is rotating with an angularvelocity of w0when it is placed on the floor. Determine the time before it

starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is μk.

1 answer:

erastova [34]4 years ago

8 0

Answer:

Explanation:

solution found below

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Compare these two waves.

Anastaziya [24]

Answer:

id 272 966 3863

Pas 161005

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3 years ago

An object is traveling on a circle with a radius of 6 feet. If in 80 seconds a central angle of 9/4 radians is swept out, then f

trapecia [35]

Answer:

The angular speed of the object is 0.0281 rad/s

The linear speed of the object is 0.169 ft/s

Explanation:

Given;

radius of the circle, r = 6 ft

time of motion of the object around the circle, t = 80 s

central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad

The angular speed of the object is calculated as;

$\omega = \frac{\theta }{t} = \frac{2.25 \ rad}{80 \ s} = 0.0281 \ rad/s$

The linear speed of the object is calculated as;

v = ωr

v = 0.0281 rad/s x 6ft

v = 0.169 ft/s

8 0

3 years ago

Which, if any, of the following statements about electric field lines is/are true? A.The electric field is always perpendicular

jonny [76]

The electric field is always perpendicular to the surface outside of a conductor. TRUE

<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>

<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>

It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>

If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>

It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance

If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
<span> FALSE: being positive, it would move in the SAME direction as the field</span>ic

8 0

3 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 $t$ :

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

5 0

3 years ago

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

Ipatiy [6.2K]

Answer:

m v1 = (m + M) v2

v2 = m v1 / (m + M)

v2 = 7 * 74 / (74 + 65)

3.73 m/s

74 kg is too heavy for the cannonball (over 150 lbs)

4 0

3 years ago