Answer:
Explanation:
In case of gas , work done
W = ∫ p dV , p is pressure and dV is small change in volume
the limit of integration is from Vi to Vf .
= ∫ p dV
= ∫ p₀
dV
= p₀ 
/ ( 
)
= - 5p₀ 
Taking limit from Vi to Vf
W = - 5 p₀ ( 
) ltr- atm.
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Hope this helps! :)
Explanation:
It is given that,
Mass of the ball, m = 1 lb
Length of the string, l = r = 2 ft
Speed of motion, v = 10 ft/s
(a) The net tension in the string when the ball is at the top of the circle is given by :
F = 18 N
(b) The net tension in the string when the ball is at the bottom of the circle is given by :
F = 82 N
(c) Let h is the height where the ball at certain time from the top. So,
Since, 
Hence, this is the required solution.
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.
A.What is the effect of air resistance on the cannonball? It's a question related to science and the experiment, which may pop into your head.
