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neonofarm [45]
3 years ago
6

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe

riences a force of magnitude F. In the three scenarios, q, ????, and r are changed to some multiple of their original values. In each case, fill in the missing value on one of the variables that will result in a force between the point charges that is equal in magnitude to the original F.
Original q Q r
Scenario 1 2q ...Q 4r
Scenario 2 ...q 2Q (1/2)r
Scenario 3 (3/2)q (3/8)Q ...r
Physics
1 answer:
sammy [17]3 years ago
6 0

Answer:

1)  Q ’= 8 Q ,  2)    q ’= 16 q ,  3)   r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

      F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

                q, Q, r

Scenario 1

      q ’= 2q

       r ’= 4r

     F = k q ’Q’ / r’²

we substitute

     F = k 2q Q ’/ (4r)²

     F = k 2q Q '/ 16r²

we substitute the value of F

      k q Q / r² = k q Q '/ 8r²

       Q ’= 8 Q

Scenario 2

       Q ’= Q

       r ’= 4r

we substitute

      F = k q ’Q / 16r²

      k q Q / r² = k q’ Q / 16 r²

      q ’= 16 q

Scenario 3

      q ’= 3/2 q

      Q ’= ⅜ Q

we substitute

        k q Q r² = k (3/2 q) (⅜ Q) / r’²

        r’² = 9/16 r²

        r ’= ¾ r

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german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

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Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

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Explanation:

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Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

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Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

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Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

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3 years ago
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