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anyanavicka [17]
3 years ago
6

A man stands on a platform that is rotating at 3.8 rpm; his arms are outstretched and he holds a brick in each hand. The rotatio

nal inertia of the system consisting of the man, bricks, and platform about the central axis of the platform is 4.65 kg-m². By moving the bricks the man decreases the rotational inertia of the system to 3.4 kg-m².
(a) what is the resulting angular speed of the platform and
(b) what is the ratio of the new kinetic energy of the system to the original kinetic energy?

Physics
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

a) 5.197rev/s

b) Kf/Ki =2.28

Explanation:

a) Angular momentum of the system L = Iw

ButLi=Lf

Kiwi =Ifwf

wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s

b)Kinetic energy KE= 0.5Iw^2

Ki = 0.5Iiwi^2

Kf=0.5Ifwf^2

Kf/Ki = Ifwf/Iiwi

Kf/Ki = (4.65/3.4))(5.197/3.8)

Kf/Ki = 1.22(1.368)^2

Kf/Ki = 2.28

kiruha [24]3 years ago
7 0

Answer:

A. 5.2 rpm.

B. Kf/Ki = 1.37.

Explanation:

Below is an attachment containing the solution.

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Leokris [45]

Answer:

B

Explanation:

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How many earths could fit inside jupiter
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1,300 or more.

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6 0
1 year ago
Two equal resistors are connected in series with a 1.50V battery. In order to keep the current at 0.030 A, the resistors much ea
AlexFokin [52]

Answer: 25 Ohms

Explanation:

From this question, the following parameters are given:

Voltage V = 1.5 v

Current I = 0.03A

From Ohm's law;

V = IR

Where R = resultant resistance of the two resistors.

Substitute V and I into the formula and make resultant R the subject of formula.

1.5 = 0.03 × R

R = 1.5/0.03

R = 50 Ohms

From the question, it is given that Thr two equal resistors are connected in series.

R = R1 + R2

But R1 = R2

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R1 = R2 = 25 Ohms

Therefore, the resistors must each have a value of 25 Ohms

4 0
3 years ago
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