Answer:
D. Friction and air resistance created heat on his trip up the hill.
Explanation:
Energy transformation from one form to another is not 100% efficient. This is the postulate of the first law of thermodynamics.
Most of the energy transformation is not purely 100%.
When energy is transformed, some are usually wasted.
- In this case, in moving from bottom up, Superman produced some heat and encountered air resistance.
- To reach the top, he must have overcome the resistance and produce enough heat to power him through.
- This reduces the amount of potential energy that should have been the same as the kinetic energy down below.
Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
So here, you're looking for distance. The formula is D=vt+1/2at^2.
Lets plug in the informations.
10 m/s is our v (initial velocity)
5 second is out t(time)
3.2 m/s is our a(acceleration)
10m/s(5)+1/2(3.2m/s+5^2)
50m/s+1/2(28.2)
50m/s+14.1
Answer =64.1 m
Glad to help you out buddy. Let me know if you need help.
Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
Answer:
f = 347.08 N
Explanation:
The frictional force exerted by the floor on the refrigerator is given as follows:
where,
f = frictional force = ?
μ = coefficient of static friction = 0.58
W = Weight of refrigerator = mg
m = mass of refrigerator = 61 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>f = 347.08 N</u>
