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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 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
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Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

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(A) we need to calculate the work done by friction in crossing the patch

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W=-\mu mg\times l

Put the value into the formula

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W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

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K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

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\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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