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STALIN [3.7K]
3 years ago
9

A skater makes one revolution per second, f = w/(2p) = 1, with her arms and leg stretched (left). By lowering her leg and bringi

ng her arms close to her chest (right) she reduces her moment of inertia from I to I′ = 0.4I. How many revolutions per second, f′ = w¢/(2p), is she making now? Show all your work.
Physics
2 answers:
mihalych1998 [28]3 years ago
8 0

Answer:

f'=w'/2p=2.49rev/s

Explanation:

The angular momentum L must be equal for both cases.

Hence you have that

L before = L after

L=I\omega

\omega=2\pi f

I = moment of inertia

w =angular momentum

Hence

L=L'\\I\omega =I'\omega'\\\omega'=\frac{I\omega}{I'}=\frac{I*2\pi *1}{0.4I}=15.70\frac{rad}{s}

f'=w'/2p=2.49rev/s

hope this helps!!

Firlakuza [10]3 years ago
4 0

Answer:

Answer is given in the attachment.

Explanation:

Download pdf
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3 years ago
Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
Virty [35]

Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
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Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

7 0
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