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IceJOKER [234]
3 years ago
11

A cargo elevator on Earth (where g = 10 m/s2) lifts 3000 kg upwards by 20 m. 720 kJ of electrical energy is used up in the proce

ss. What is the efficiency of this process?
Physics
1 answer:
MakcuM [25]3 years ago
8 0

Answer: 83%

Explanation:

Efficiency of the process = work output/work input × 100%

Work input is the energy used up in the process = 720,000Joules

Work output = Force × distance

= (3000×10)× 20

= 600000 Joules

Efficiency= 600000/720000 × 100

= 0.83×100

= 83%

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One side of the moon always faces Earth because the time it takes the moon to spin on its axis is blank the time it takes the mo
Kamila [148]
The word to fill in the blank is "equal". Because the time taken to rotate (spin on its axis) is equal to the time of revolution (going around the earth), this means that both have the same rate of angular rotation. So for every bit that the moon goes around its orbit around earth, the moon itself rotates accordingly to present the exact same side to earth.
6 0
3 years ago
A train travels 8.81 m/s in a -51.0° direction.
Amiraneli [1.4K]

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

  • θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

#SPJ1

8 0
1 year ago
The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k
Brrunno [24]

Answer:

The gravitational force changing velocity is

\frac{dF}{dt}=-8\frac{N}{s}

Explanation:

The expression for the gravitational force is

F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}

Differentiate the above equation

\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}

The velocity is the distance in at time so

V=\frac{dr}{dt}=0.4 \frac{km}{s}

\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

\frac{dF}{dt}=-8\frac{N}{s}

8 0
3 years ago
For work to be done on the object the object has to
Nata [24]
Gain energy

good luck!
3 0
3 years ago
Read 2 more answers
A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from
Alla [95]

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

6 0
3 years ago
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