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sdas [7]
3 years ago
15

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

m of the stone about the center of the circle is:
Physics
1 answer:
Illusion [34]3 years ago
3 0
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
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What is the net force acting upon this object? *
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.. .. .. .. .. .. .. ..

4 0
3 years ago
Read 2 more answers
1. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600 Vo
JulsSmile [24]

Answer:

\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s

Explanation:

First, we will calculate the inductance of the solenoid by using the following formula:

L = \frac{\mu_o AN^2}{l}

where,

L = self-inductance of solenoid = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

A = Cross-sectional area = 30 cm² = 3 x 10⁻³ m²

N  = No. of turns = 2000

l = length = 65 cm = 0.65 m

Therefore,

L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L =  0.0232\ H

Now, we will use Faraday's law to calculate the rate of change of current:

emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\  \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s

5 0
3 years ago
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