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ioda
3 years ago
11

A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion

?
Physics
2 answers:
Oksana_A [137]3 years ago
7 0

Answer:

0.472rad/s

Explanation:

Angular velocity = 2πf where f = frequency and frequency is the number of revolution per second and 2π represent a cycle of revolution. The mass of the body was 5kg, the time taken to complete a cycle was 13.3 s.

Frequency = 1/period where period is the time it takes to complete a revolution.

F = 1/13.3 = 0.075hz

Angular velocity = 2* 3.142* 0.075 = 0.472rad/s

Tcecarenko [31]3 years ago
6 0

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

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A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a
lubasha [3.4K]
The angular speed can be solve using the formula:
w = v / r
where w is the angular speed
v is the linear velocity
r is the radius of the object

w = ( 5 m / s ) / ( 5 cm ) ( 1 m / 100 cm )
w = 100 per second
8 0
3 years ago
Read 2 more answers
Would the forces acting on the sky diver be balanced or unbalanced? Explain your answer.
docker41 [41]

Answer:

your mom is balanced!!!

4 0
3 years ago
A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample
Degger [83]

It must be the same force, because when it is applied, it is 'distributed' equally in each of the points (either a tendon or a wire or a rope or something similar). The 128N resistance should be preserved for tendons that have the same diameter.

Therefore the correct answer is A.

3 0
3 years ago
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