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3 years ago

7

Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The

student’s speed is meters/second

1 answer:

defon3 years ago

3 0

225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s

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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

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Explanation:

It is known that relation between torque and angular acceleration is as follows.

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and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

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as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

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Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

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3 years ago

"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of

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2 years ago

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Answer:

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B. $Q=2112\ J$

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D. $W=0\ J$

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Given:

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initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

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A.

<u>Work done by the gas during the initial isobaric expansion:</u>

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$W=2\times 10^5\times (0.006-0.003)$

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C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

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$\Delta U=1512\ J$

B.

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$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

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$W=0\ J$

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2 years ago

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Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If, V(0) = 0, V(1) = 2

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Now Check V(t) = A× t + B × t^²

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Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

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