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3 years ago

11

Two hockey players , big Jim and little Tim collide head on and get tangled while going for the puck 65kg time was traveling at

9.5m/s while while 105 kg joke was moving at 5.2 m/s what is big joes momentum before the collision

1 answer:

12345 [234]3 years ago

7 0

I will assume that big Joe is big Jim. The equation for the momentum is p=m*v, where m is the mass of the body and v is the velocity. Big Joe has a mass m=105 kg and speed v=5.2 m/s. When we input the numbers:

p=105*5.2=546 kg*(m/s).

So big Joe's momentum before the collision is p=546 kg*(m/s).

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How does a television have a negative effect on the environment

max2010maxim [7]

Answer:

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Explanation:

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3 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

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3 years ago

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Ipatiy [6.2K]

Answer:

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Explanation:

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Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.

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What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

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Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

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3 years ago