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Stolb23 [73]
2 years ago
15

I need help with some physics stuff​ HELP

Physics
1 answer:
Airida [17]2 years ago
4 0

Answer:

You just double the number of boxes it goes higher. Then triple the number for the second question. But don't change the time.

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What is work done in physics and how can it be calculated
shusha [124]
Work is done when a force is applied to an object moves that object. the work is calculated by multiplying the force by the amount of movement of an object
8 0
3 years ago
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Bob is pulling a 30kg filing cabinet with a force of 200n , but the filing cabinet refuses to move. the coefficient of static fr
puteri [66]
The cabinet is being pulled with 200N and is being rested by a force equal to 200N. That is why it is not being moved. 


<span>Although the force of static friction can equal Fk=µs*F=m*g*µs=(30kg)*(9.8m/s^2)*(0.80)=235 N. It is not resisting the 200N force with 235N. Imagine if you pushed something with 200N and it pushed you back with 235N, especially a cabinet. You would think that the cabinet was alive.</span>
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4 years ago
A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction
marissa [1.9K]

Answer:

Part a: <em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b: <em>The total amount of energy transfer by the heat is 54.81 kJ</em>

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of CO_2 is given as

M=44 kg/kmol

the gas constant is given as

R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K

Volume of the tank is given as

V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3

Final mass is given as

m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg

Mass of the CO2 moved to the cylinder

m=m_1-m_3\\m=3-2=1 kg

The initial mass in the cylinder is given as

m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg

The mass after the process is

m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\

Now the volume 2 of the cylinder is given as

V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3

Part a:

So the Work done is given as

W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ

<em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b:

The total energy transfer by heat is given as

Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ

As the temperature is constant thus change in internal energy is 0.

<em>The total amount of energy transfer by the heat is 54.81 kJ</em>

7 0
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All three of those are correct.
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4 years ago
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Mashcka [7]
I believe its  A . green 
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3 years ago
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