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iVinArrow [24]
2 years ago
3

Define a function computegasvolume that returns the volume of a gas given parameters pressure, temperature, and moles. use the g

as equation pv = nrt, where p is pressure in pascals, v is volume in cubic meters, n is number of moles, r is the gas constant 8.3144621 ( j / (mol*k)), and t is temperature in kelvin. all parameter types and the return type are double.
Physics
2 answers:
raketka [301]2 years ago
8 0

Answer:

Incomplete question, we are given the starter code below

Explanation:

The starter code given

gas_const = 8.3144621

def compute_gas_volume((gas_pressure,,gas_temperature,gas_moles):

gas_pressure = 100.0

gas_moles = 1.0

gas_temperature = 273.0

gas_volume = 0.0

gas_volume = compute_gas_volume(gas_pressure, gas_temperature, gas_moles)

print('Gas volume:', gas_volume, 'm^3')

Then, we will want to print

pv = nrt

Gas volume is

v = nrt/p

v = n×r×t ÷ p

So to compute v, we use

public static void main(String args[]){

final Double R = 8.3145;

Double pressure;

Double volume;

Double n;

Double temperature;

System.out.println("What is the number of moles?

n = scan.nextDouble();

System.out.println("What is the tempertaure in K?

temperature = scan.nextDouble();

System.out.println("What is the pressure in atm?

pressure = scan.nextDouble();

volume=(n*R*temperature)/pressure

System.out.println("The volume is " + volume + " m^3 or L");

Since we are already given the value of the parameter. Then, we can write the code as follow

gas_pressure = 100.0

gas_moles = 1.0

gas_temperature = 273.0

gas_volume = 0.0

gas_volume =

GAS_CONST = 8.3144621

defcompute_gas_volume(gas_pressure, gas_temperature, gas_moles):

return gas_moles * GAS_CONST * gas_temperature / gas_pressure

user100 [1]2 years ago
5 0

Answer:

Refer below.

Explanation:

A function can be characterized as a connection wherein one thing is reliant on another for its worth.

Clarification: Given R = 8.314J/mol*k

PV = nRT

V = nRT/P

V = 8.314RT/P (cm^3)

Volume of gas =(( 8.314 * R* T)/P ) cm3

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Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

r = 80\;\text{cm} = 0.80\;\text{m}.

What's the net force on the object?

The object is in a circular motion. As a result,

\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r},

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  • m is the mass of the object,
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For this object,

\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}.

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

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What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

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All three values in this question are given with two sig. fig. Round the value of T to the same number of significant figures.

4 0
2 years ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

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