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3 years ago

What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds?

Physics

1 answer:

slava [35]3 years ago

7 0

Answer:

The tangential velocity of a rotating object is:

v = r*w

where r is the radius, and w is the angular velocity.

w = 2*pi*f

where f is the frequency.

We know that the record plater does 11 revolutions in 20 seconds, then it does:

11 rev/20s = 0.55 rev/s = f

then we have:

w = 2*pi*0.55 s^-1 = 2*3.14*0.55 s^-1 = 3.454 s^-1

The radius of a record player is really variable, it is around 10 inches, so i will use r = 10in, which is the rotating part of the record player.

then the tangential velocity is:

v = 10in*3.454 s^-1 = 34.54 in/s

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$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.

In this problem:

$q_1 =6 \mu C=6 \cdot 10^{-6}C$

$q_2=2 \mu C=2 \cdot 10^{-6}C$

$r=0.1 m$

Substituting into the equation, we find

$F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N$

(b) direction of particle q2

Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.

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3 years ago

Read 2 more answers

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Answer:

$4.81*10^{29}J$

Explanation:

Since the formula for kinetic energy of an object is:

$E_k = \frac{mv^2}{2}$

Where m is the mass of the object and v is the speed. We can substitute m = $3*10^{21}$ kg and v = 17900m/s:

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2 years ago

A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot

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Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º, T total=2*Voy/g

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Velocity in the Y axis:

Voy=Vo*sinα

Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ; g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

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α = 90º, sinα=1

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3 years ago

Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry

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Answer is given below

Explanation:

This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle
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3 years ago

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The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

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For the west or W direction, we use the equations:
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d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
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t = 18.33 seconds
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Thus,
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