Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
Answer:
Hindi ko alma yam among twang yan
Explanation:
aorry
Answer:
Farm = 98.1 [N]
Explanation:
To solve this problem we must draw the respective free body diagram, with the forces acting on the monkey. An analysis of the sums on the y-axis must be performed, in this axis the weight is acting down and the forces of both arms pulling up.
Weight is defined as the product of mass by gravitational acceleration.
W = m*g
where:
m = mass = 20 [kg]
g = gravity acceleration = 9.81 [m/s²]
W = 196.2 [N] (units of Newtons)
As this force points down, the force of both arms must go up, therefore each arm exerts a force of:
Farm = 196.2 / 2
Farm = 98.1 [N]
Carbohydrates <span>carbohydrates are repeating sugar units. They are the only ones that are repeating sugar units</span>
