Answer:
5 watts
power = work done ÷ time taken
10 j ÷ 2 = 5w
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
Answer:
Explanation:
We shall apply conservation of mechanical energy law to solve the problem .
loss of height = L ( 1 - cos 37 ) where L is length of rope
loss of potential energy at the bottom = gain of kinetic energy .
mg L ( 1 - cos 37 ) = 1/2 m v² where v is velocity at the bottom
v² = 2 L g ( 1 - cos 37 )
= 2 x 10 x 9.8 ( 1 - cos 37 )
= 39.46
v = 6.28 m /s
ºC = K - 273
ºC = 600 - 273
T = 327ºC
Answer C
hope this helps!