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brilliants [131]
3 years ago
12

The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what

is its speed at the top of 25 m peak(Point s
Physics
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

22 m/s

Explanation:

PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf

−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s

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Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
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The weight of a girl with a mass of 40.0 kg is_N
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Explanation:

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2 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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PLEASE HELP! It’s urgent... and please show your work!!
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Answer:

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