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GrogVix [38]
3 years ago
8

Calculate the average power required to lift a 750 N object a vertical distance of 10 meters in 4.0 seconds.

Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:

500

Explanation:

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A certain 100W light bulb has an efficiency of 95%. How much thermal energy will this light bulb add to the inside of a room in
Usimov [2.4K]
Since the bulb consumes 100 watts of power and its efficiency is 95%,
it generates 95 watts of light energy and 5 watts of heat energy whenever
it's turned on.

5 watts means  5 joules of energy per second.

(2.5 hours) x (3,600 seconds/hour) =  9,000 seconds

(9,000 seconds) x (5 joules/second)  =  45,000 joules of heat in 2.5 hours

7 0
3 years ago
Suppose your car is on a 5% grade, meaning that for every 100 m you travel along the road you raise or lower only 5 m in elevati
kvv77 [185]

Answer:

734.215N

Explanation:

First we calculate the angle that corresponds to a 5% slope using the Tan-1 function

\beta = tan-1(5%)=2.86

then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)

Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N

8 0
3 years ago
Read 2 more answers
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in
insens350 [35]

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

5 0
3 years ago
A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

5 0
3 years ago
Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega
devlian [24]

Answer: its negative

Explanation: becase it is

5 0
3 years ago
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