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Anika [276]
3 years ago
13

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s. The circle is parallel to the xy pl

ane and is centered on the z axis, 0.75 m from the origin. The magnitude of its angular momentum around the origin is:
Physics
2 answers:
Lynna [10]3 years ago
6 0
It's 60.... i have used the formula
L=r^2mw
and didn't use the value 0.75...
m_a_m_a [10]3 years ago
3 0
As we know that
L=r^2mw
=(0.75)^2 ×2×12
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Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
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Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

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A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
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Answer:

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Given the following data;

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344a = 858.49 - 210.25

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V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

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