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timama [110]
3 years ago
5

An athlete does one push-up. In the process, she moves half of her body weight, 250 newtons, a distance of 20 centimeters. This

distance is the distance her gravity moves when she fully extends her arms. How much work did she do after one push-up?
Physics
1 answer:
s2008m [1.1K]3 years ago
6 0
We know that an athlete moves half of her body weight 250 N a distance of 20 cm.
So:  F = 250 N,  d = 20 cm = 0.2 m
The work formula:
W = F * d
W = 250 N * 0.2 m
W = 50 J
Answer: Her work after one push-up is 50 J.
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The next town to where I live is 25 km away. On a good day, I can make it there in about 3/4 of an hour.

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Using diagram differentiate between solenoid and a toroid
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The Toroid is form when you have wound conductor around circular body. In this case you have magnatic field inside the core but you dont have any poles because circular body dont have ends. This can be used where you want minimum flux leakage and dont need magnatic poles. i.e. toroidal inductor, toroidal transformer.


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16. Kinematic equations can only be used if...​
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A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

3 0
3 years ago
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