Answer:
22.65 rad/s
Explanation:
m = 7 kg, r = 0.21 m, F = 44 N, d = 0.9 m
Moment of inertia, I = 1/2 mr^2 = 0.5 x 7 x 0.21 x 0.21 = 0.15435 kg m^2
Work done, W = F x d = 44 x 0.9 = 39.6 J
according to the work energy theorem
Change in the kinetic energy of rotation = Work done
1/ 2 x I x ω^2 = W
0.5 x 0.15435 x ω^2 = 39.6
ω^2 = 513.12
ω = 22.65 rad/s
Here we can use energy conservation
As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc
Now we have
now we know that work done is product of force and displacement
so here we have
now for moment of inertia of the disc we will have
now from above equation we will have
Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.
-4.71 m/s
Given:
y₀ = 1.13 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)
v = -4.71 m/s
The answer is A). Moving from A to C the temperature and the kinetic energy increases.