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Art [367]
3 years ago
13

5. Two charged particles are separated by a distance of 12 meters. The Coulomb force between them is 20 N. What will the Coulomb

force be if the same particles are
separated by a distance of 6 meters?
O A 80 N
OB. 40 N
O C. 10N
O D5N
Physics
1 answer:
Leni [432]3 years ago
8 0

Answer:

A) 80 N

Explanation:

The closer the particles get, the stronger the Coulomb force, which elongates choices C and D. The Coulomb force is inversely proportional to the distance squared. If the distance is cut in half, the force is multiplied by the reciprocal of (1/2)^2, which is 4. Multiplying it out, 20 times 4 is 80 N.

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pickupchik [31]
A balloon expanding would count as work because it is moving and none of the others are. However, I think that James pedaling also works because at least he is using his leg muscles even though he's not moving. He is working because he is exercising himself and that count as working. I would chose James over all of them.
7 0
3 years ago
Read 2 more answers
While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau
eduard

Answer:

The speed is   v =8.17 m/s

Explanation:

From the question we are told that

      The angle of slant is  \theta = 37.0^o

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       The mass of the toolbox is m = \frac{92}{9.8} = 9.286kg

       The start point is  d = 4.25m from lower edge of roof

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Generally the net work done on this tool box can be mathematically represented as

      Net \ work done  =  Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction

The workdone due to weigh is  =    mgsin \theta * d

 The workdone due to friction is  = F_f \ cos\theta  *   d

Substituting this into the equation for net workdone  

                 W_{net} = mgsin\theta  * d + F_f  \ cos \theta *d

      Substituting values

                  W_{net}  =  92 * sin (37)  * 4.25 + 22 cos (37) * 4.25

                          = 309.98 J

 According to work energy theorem

             W_{net} = \Delta Kinetic \ Energy

              W_{net} = \frac{1}{2} m (v - u)^2

From the question we are told that it started from rest so  u = 0 m/s

              W_{net} = \frac{1}{2} * m v^2

Making v the subject

               v = \sqrt{\frac{2 W_{net}}{m} }

Substituting value

              v = \sqrt{\frac{2 * 309.98}{9.286} }

             v =8.17 m/s

6 0
3 years ago
If a ball is dropped from a height​ (H) its velocity will increase until it hits the ground​ (assuming that aerodynamic drag due
mario62 [17]

Answer:

9.801 m/s²

Explanation:

t = Time taken

u = Initial velocity

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Acceleration of the ball is 9.801 m/s²

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Verizon [17]
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3 years ago
WHAT SHOULD I NAME MY DEAD RAT?
olga_2 [115]

Answer:

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Explanation:

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