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3 years ago

A clock moves past you at a speed of 0.9 c. How much time passes for you for each second that elapses on the moving clock?

Physics

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

Explanation:

Time dilation formula is

T = T₀ / √ 1-v²/c²

T₀ is time elapsed in moving reference , T time elapsed in stationary reference.

Here T₀ = 1 second

T = 1/√ 1-0.9² = 1/.4358 = 2.3 second

So 2.3 second will pass for each second on moving reference.

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A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

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Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

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The trachea is a tube that carries air inside the lungs.

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The air that flows over the top part of an airplane's wing moves faster than the air that flows across the bottom. This faster m

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As an airplane moves through the air, its wings cause changes in the speed and pressure of the air moving past them. These changes result in the upward force called lift.

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A wing is shaped and tilted so the air moving over it moves faster than the air moving under it. As air speeds up, its pressure goes down. So the faster-moving air above exerts less pressure on the wing than the slower-moving air below. The result is an upward push on the wing—lift!

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Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve

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Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

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Explanation:

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