The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
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The answer is A bc I did the quiz and I got it right
Answer:
a = 0.55 m / s²
Explanation:
The centripetal acceleration is given by the relation
a = v² / r
angular and linear velocities are related
v = w r
we substitute
a = w² r
In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system
w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s
let's calculate
a = 0.105² 50.0
a = 0.55 m / s²
Answer:
<u>The car's fast. The ground isn't moving.</u>
Hope this helped! :D
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
