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Trava [24]
3 years ago
15

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

xle through its center, but at one point on its equator, it is scraping against metal, resulting in a friction force of 0.0200 N at that point.
(a) Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
(b) How long will it take to decrease its rotational speed by 22.0 rad/s?

Physics
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer:

Answer is in the following attachment.

                                                                                                                                                                                                 

Explanation:

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i
drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = \frac{28}{10} × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

8 0
3 years ago
Is this right or they wrong definitions which ones are the right ones someone !!!!!
Alex73 [517]

Answer:

They are right.

Explanation:

4 0
3 years ago
Read 2 more answers
A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati
Ainat [17]

Answer:

The asteroid's acceleration at this point is 2.71\ m/s^2

Explanation:

The equation that governs the trajectory of asteroid is given by :

x=6.5t-2.3t^3

The velocity of asteroid is given by :

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

6.5-6.9t^2=0\\\\t=0.971\ s                      

Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t                        

Put t = 0.971 s

a=-13.8\times 0.197\\\\a=-2.71\ m/s^2

So, the asteroid's acceleration at this point is 2.71\ m/s^2 and it is decelerating.

6 0
3 years ago
A man, a distance d=3~\text{m}d=3 m from a target, throws a ball at an angle \theta= 70^\circθ=70 ​∘ ​​ above the horizontal. If
lbvjy [14]

Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

4 0
3 years ago
PLEASE HELP!!!! Thank you!
V125BC [204]

it would be a chain reaction and a wave

6 0
4 years ago
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