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Aleks [24]
3 years ago
5

A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from its left end. If a force of 50N is appl

ied on it's right end, how much force would need to be applied to the left end?
100N
25N
200N
50N
​
Physics
1 answer:
grigory [225]3 years ago
6 0
The force that would be apply is
100N
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If the needle in the galvanometer is going up, what can you conclude about the motion of the magnet in the diagram?
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Answer:

t is not moving.

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Answer is: It must be moving right

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Why machines are never 100 percent efficient
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Machines are never 100% efficient because they are made by humans and we make errors and nothing we make is perfect, that's why the machines we build are not 100% accurate or efficient.
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Calculate the period (T) of uniform circular motion if the velocity is 40.0 m/s and centripetal acceleration is 20.0 m/s2.
kirill [66]
T is the time for a whole round.

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And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s
3 0
3 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
monitta

Answer:

force F = 1.66 × 10^{-13} N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r²      ......................1

put here value and we get

force F = 9 × 10^{9} × \frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}    

force F = 1.66 × 10^{-13} N

4 0
3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
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Answer:1.71 m/s

Explanation:

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mass of Susan m=12 kg

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coefficient of Friction \mu =0.18

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F_x=T\cos \theta -f_r

where f_r=friction\ Force  

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since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

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Work done by applied Force is equal to change to kinetic Energy

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