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Zarrin [17]
2 years ago
13

A 1.00 x 104 W electric motor is used to lift a

Physics
1 answer:
Alexus [3.1K]2 years ago
4 0

Answer: 52%

Explanation:

1W = 1 J/s

motor input is 10000 J/s

Potential energy change

PE = 955(9.81)(25.0) = 234,213.75

power needed to change the PE in that time

P = 234,213.75/ 45 = 5,204.75 Watts

motor is 5204.75 / 10000 = 0.520475 or 52% efficient

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Superman does an exhibition run at a track meet. When he runs the 200 m
Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered  = 200m

Acceleration  = 10m/s²

Unknown:

Time taken to cover the track  = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

       S  = ut + \frac{1}{2} at²  

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

 So;

     S  =  \frac{1}{2} at²  

        200  =  \frac{1}{2} x 10 x t²  

          200  = 5t²  

            t²  = 40

            t  = 6.32s

5 0
2 years ago
What is the height of the image? Round the answer
Maru [420]
It’s a concave mirror
8 0
2 years ago
To shoot a swimming fish when an intense light beam from a laser gun you should aim
Ivan

Answer

aim directly at the image

Explanation

the light from the laser beam will also bend when it hits the air water interface , so aim directly at the fish

5 0
3 years ago
Light waves are not mechanical waves. The Sun transmits light waves to Earth through __________ _________. *
Natasha2012 [34]
Electromagnetic waves?
3 0
3 years ago
Read 2 more answers
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
3 years ago
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