Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
5.5 s
Explanation:
The time it takes for the ball to reach its maximum height can be calculated using

since
at the top of its trajectory. Plugging in the numbers,

Answer:
η = 40 %
Explanation:
Given that
Qa ,Heat addition= 1000 J
Qr,Heat rejection= 600 J
Work done ,W= 400 J
We know that ,efficiency of a engine given as

Now by putting the values in the above equation ,then we get

η = 0.4
The efficiency in percentage is given as
η = 0.4 x 100 %
η = 40 %
Therefore the answer will be 40%.