Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2
I believe it is called centripetal force <span />
Answer:37 J
Explanation:
Given
Step :1
Heat added Q=44 J
Work done=-20 J
![\Delta E_1=Q+W=44-20=24 J](https://tex.z-dn.net/?f=%5CDelta%20E_1%3DQ%2BW%3D44-20%3D24%20J)
Step :2
Heat added Q=-61 J
work done ![W_2](https://tex.z-dn.net/?f=W_2)
![\Delta E_2=Q+W_2](https://tex.z-dn.net/?f=%5CDelta%20E_2%3DQ%2BW_2)
![\Delta E_2=61+W_2](https://tex.z-dn.net/?f=%5CDelta%20E_2%3D61%2BW_2)
![\Delta E_1+\Delta E_2=0](https://tex.z-dn.net/?f=%5CDelta%20E_1%2B%5CDelta%20E_2%3D0)
as the process is cyclic
![44-20-61+W_2=0](https://tex.z-dn.net/?f=44-20-61%2BW_2%3D0)
![W_2=37 J](https://tex.z-dn.net/?f=W_2%3D37%20J)
work done in compression is 37 J
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479