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alekssr [168]
3 years ago
6

A solid sphere is placed on a frictionless floor in a very long corridor and is given a quick push so that it begins to slide, w

ithout rotating, along the corridor. How would the angular speed of the sphere be changing if the floor were not frictionless?
A.) It would remain zero because the net torque is still zero.
B.) It would remain zero because the angular momentum is conserved.
C.) It would be increasing until the slipping between the sphere and the floor stops.
D.) It would be increasing until the linear and the angular speeds become equal.
E.) It would be increasing until the translational motion stops.

Physics
1 answer:
prohojiy [21]3 years ago
8 0
I think it’s D I hope it helps
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Answer:

Fundamental frequency in the string will be 25 Hz

Explanation:

We have given length of the string L = 1.2 m

Speed of the wave on the string v = 60 m/sec

We have to find the fundamental frequency

Fundamental frequency in the string is equal to f=\frac{v}{2L}, here v is velocity on the string and L is the length of the string

So frequency will be equal to f=\frac{v}{2L}=\frac{60}{2\times 1.2}=25Hz

So fundamental frequency will be 25 Hz

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What happens to gravitational potential energy as a roller coaster moves down a hill?
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viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

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at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

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now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

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3 years ago
Madison was driving at 40 mph and went 80 miles. How long did it take madison?
monitta
v=40\ mph\\\\s=80\ miles\\\\t=?\\--------------\\v=\frac{s}{t}\to vt=s\to t=\frac{s}{v}\\--------------\\t=\frac{80\ miles}{40\ mph}=2\ h\leftarrow Answer
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Answer:

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