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Feliz [49]
3 years ago
9

Two otherwise identical, small conducting spheres have charges +7.0 µC and −3.0 µC. When placed a distance r apart, each experie

nces an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find the magnitude of the new force on each sphere.
Physics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

0.7 N

Explanation:

q1 = 7 micro coulomb

q2 = - 3 micro coulomb

distance = r

Force, F = 3.7 N

Let the new force be F'.

By use of Coulomb's law

F=\frac{Kq_{1}q_{2}}{r^{2}}

3.7=\frac{K\times 7\times 10^{-6}\times 3\times 10^{-6}}{r^{2}}        .... (1)

Now touch the spheres,

q1 = q2 = (7 - 3 )/ 2 = 2 x 10^-6 C

Now

F'=\frac{K\times 2\times 10^{-6}\times 2\times 10^{-6}}{r^{2}}     .... (2)

Dividing equation (2) by equation (1), we get

\frac{F'}{3.7}=\frac{4}{21}

F' = 0.7 N

Thus, the new force is 0.7 N.

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Helium

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Helium is the least reactive element, since it is a noble gas with the smallest amount of valence rings.

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Why does a third class lever cannot magnify force?​
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A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of
dem82 [27]

Answer:

v=v0 - gt

Explanation:

The equation for velocity is

v=v0 - gt

where v0=14m/s, g=10m/s^2.

in 1 second:

v=14-10=4m/s

it is positive so direction is upwards

in 2 seconds:

v=14-20=-6m/s

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5 0
3 years ago
A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.
iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 32}

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

J=m(v-u)

J=1.2\times (-25.04-10)

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

J=\dfrac{F}{t}

F=J\times t

F=42.048\times 0.02

F = 0.8409 N

Hence, this is the required solution.

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3 years ago
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