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crimeas [40]
3 years ago
5

A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

olid disc with a radius of 2.50 m and a mass of 10.0 kg is spinning at 3500. rpm. The smaller disc is carefully pressed against the larger disc (flat side to flat side) so that they spin together without slipping. What is the angular velocity of the large disc
Physics
1 answer:
drek231 [11]3 years ago
4 0

Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger  disc I₁ = 1/2 m r²  , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller  disc I₂ = 1/2 m r²  , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂  = ( I₁ +I₂) ω  , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

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A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha
ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, L=400\ \mu H=400\times 10^{-6}\ H

Capacitance of parallel LCR circuit, C=0.3\ \mu F=0.3\times 10^{-6}\ F

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

f=\dfrac{1}{2\pi \sqrt{LC} }

f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }

f=14528.79\ Hz

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0
3 years ago
ONLY TWO ANSWER CHOICES!!
Otrada [13]
A: Buoyant force is equal to the weight
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3 years ago
A car travels west for 240 km in 4 h. what is the car's velocity?
nalin [4]
The velocity is 60 because you divide your distance by your time (240÷4=60)
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3 years ago
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What's the word used to describe the weight of precious stones.​
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2 years ago
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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
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