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skad [1K]
2 years ago
8

Give at least 4 human activities which demonstrate the involvement of inertial frame of reference​

Physics
1 answer:
Sergio039 [100]2 years ago
7 0
One clearcut example of an inertial reference frame is an isolated spaceship, far, far away from the Earth, the Sun, the Milky Way Galaxy, and all other massive objects. Fred places a blue ball into a claw at the left end of the ship, and red ball into a claw at the right end of the ship.
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In which of these cases is a waiter doing work on the object
Gnom [1K]
Section 2 is right,, i think. good luck
8 0
2 years ago
What is the use of an inclined plane​
loris [4]

Answer:

Hey there!

Inclined planes are used to lift heavy objects to higher places.

Hope this helps :)

5 0
3 years ago
Read 2 more answers
In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation
mr_godi [17]

Answer:

the equilibrium wage rate is 10  and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

W_{Demand} = W_{Supply}

15 - (1/200) L = 5 + (1/200) L

15 - 5 =  (1/200) L +  (1/200) L

10 = (2/200) L

(10*200)/2 = L

1000 = L

Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

W = 15 - (1/200) 1000

W = 10

Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

7 0
3 years ago
Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state
topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

<u>Given  </u>

<u><em>Process 1 ---> 2 </em></u>

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

Volume of point (1) V1   = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant  

<u>Process 2 ---> 3 </u>

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

<u>Process 3 ---> 1 </u>

The relation of the process PV = constant  

<u>Required  </u>

Sketch the processes on the PV coordinates

The work for each process in kJ  

<u>Solution  </u>

The work is defined by  

W=\int\limits^a_b {x} \, dx

<em>a=V2</em>

<em>b=V1</em>

<em>x=P</em>

<em>dx=dV</em>

<u>Process 1 ---> 2  </u>

P3 = P4 = 5 bar  

W=\int\limits^a_b {x} \, dx

<em>a=V3</em>

<em>b=V2</em>

<em>x=4</em>

<em>dx=dV</em>

putting the value of a, b, x, dx in above integral

W=400 kJ

<u>Process 2 ---> 3 </u>

V = constant Then there is no change in the volume,hence W = 0 kJ  

<u>Process 3 ---> 1  </u>

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1  

 W=\int\limits^a_b {x} \, dx

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

  = -160.944 KJ

5 0
3 years ago
Calculate the pressure in SI unit when a force of 100 dyne acts on area of 0.02cm²​
Andru [333]

Answer:

500 Pa

Explanation:

Convert given units to SI:

100 dyne = 0.001 N

0.02 cm² = 2×10⁻⁶ m²

Pressure = force / area

P = 0.001 N / (2×10⁻⁶ m²)

P = 500 Pa

7 0
3 years ago
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