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kiruha [24]
3 years ago
11

Two small masses that are 10.0 cm apart attract each other with a force of 10.0 N. When they are 5.0 cm apart, these masses will

attract each other with what force?
PLEASE SHOW WORK :)
Physics
1 answer:
aleksandrvk [35]3 years ago
3 0
We know, force is indirectly proportional to square of distance
F directly proportional 1/r²

You halved the distance [ from 10 to 5 ] 

New Force with 5 cm would be: = F'
F' = 1/(r/2)² = 4/r² = 4.F    [ As we have: F = 1/r²]
F' = 4F
F' = 4×10 = 40 N

In short, Your Answer would be 40 Newtons

Hope this helps!
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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

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T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

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30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

8 0
3 years ago
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m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

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f = kinetic frictional force acting on the person by the surface

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The normal force by the surface in upward direction balances the weight of the person in down direction , hence

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kinetic frictional force on the person acting is given as

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