Efficiency = (Wanted) energy out ÷ energy in × 100
Energy in = 400J
Wanted Energy out = 240J
Energy cannot be used up, only transferred, so the remaining energy is most likely to be transferred into unwanted energy (loss of energy) such as heat energy.
Efficiency = 240 ÷ 400 × 100
Efficiency = 0.6 × 100
Efficiency = 60%
Answer:
Decreases to half.
Explanation:
From the question given above, the following data were obtained:
Initial mass (m₁) = m
Initial force (F₁) = F
Initial acceleration (a₁) =?
Final mass (m₂) = ½m
Final force (F₂) = ¼F
Final acceleration (a₂) =?
Next, we shall determine a₁. This can be obtained as follow:
F₁ = m₁a₁
F = ma₁
Divide both side by m
a₁ = F / m
Next, we shall determine a₂.
F₂ = m₂a₂
¼F = ½ma₂
2F = 4ma₂
Divide both side by 4m
a₂ = 2F / 4m
a₂ = F / 2m
Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:
a₁ = F / m
a₂ = F / 2m
a₂ : a₁ = a₂ / a₁
a₂ / a₁ = F/2m ÷ F/m
a₂ / a₁ = F/2m × m/F
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
From the illustrations made above, the acceleration of the car will decrease to half the original acceleration
watts = work per second.
work is mgh = 3x10x50=1500
watts out = 1500
Watts used = 2000
eff=1500/2000=75%
Correct answer is letter B. sandstone
<span>The person is dragging
with a force of 58 lbs at an angle of 27 degrees relating to the ground. We
want to use cosine function to look for the horizontal force component. And
then we can compute for W = (Horizontal Force) x (Distance). We want the
horizontal force component since that is the component that is parallel to the
direction the cart is moving. </span><span>
(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal
force component.)
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>
