After the elevator accelerates, then it moves at a constant speed of 6.0m/s, calculate the tension inthe cable.
1 answer:
We know that
• The mass of the elevator is 5000 kg.
Let's draw a free-body diagram.
As you can observe, there are just two forces involved, the weight of the elevator and the tension force. Let's use Newton's Second Law.
But, W = mg = 5000kg*9.8m/s^2 = 49,000 N, and m = 5000 kg, a = 0 (because the speed is constant).
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Answer: The correct answer is option B.
Explanation:
Mass of the sled = 10 kg
Initial speed of the sled = 2 m/s
Kinetic energy of the sled =
Work done by the sled = 20 joules
The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.
Hence, correct answer is option B.
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
