What happens to the gravitational force between two object if the mass of the objects stay the same but distance decreases?
1 answer:
Explanation:
[ file ] Where , G is universal gravitational constant
m1 and m2 masses of object 1 and 2 respectively !!
r , is distance between the two object !!
:- If we decrease the distance between them the effect on Gravitational force will be : it increases due to the inverse proportionality between the gravitational force ( Fg ) and distance between object ( r² ) ..
..... Even it is inversely proportional to the square of the distance between object , the force will increase suddenly for a small change in r !!
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Answer:
Answered
Explanation:
a) What is the work done on the oven by the force F?
W = F * x
W = 120 N * (14.0 cos(37))
<<<< (x component)
W = 1341.71
b)
= 29.4 N
W_f= 328.72 J = 329 J
c) increase in the internal energy
U_2 = mgh
= 12*9.81*14sin(37)
= 991 J
d) the increase in oven's kinetic energy
U_1 + K_1 + W_other = U_2 + K_2
0 + 0 + (W_F - W_f ) = U_2 + K_2
1341.71 J - 329 J - 991 J = K_2
K_2 = 21.71 J
e) F - F_f = ma
(120N - 29.4N ) / 12.0kg = a
a = 7.55m/s^2
vf^2 = v0^2 + 2ax
vf^2 = 2(7.55m/s)(14.0m)
V_f = 14.5396m/s
K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J
Answer:
the location of the center of gravity for the entire body is 1.08 m
Explanation:
Given the data in the question;
w1 = 458 N, y1 = 1.34 m
w2 = 120 N, y2 = 0.766 m
w3 = 89.8 N, y2 = 0.204 m
The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.
so,
= (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )
so we substitute in our values
= (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )
= 723.9592 / 667.8
= 1.08 m
Therefore, the location of the center of gravity for the entire body is 1.08 m