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3 years ago

What is electrostatic force?

Physics

1 answer:

zalisa [80]3 years ago

5 0

"Electrostatic forces are attractive or repulsive forces between particles that are caused by their electric charges."

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A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn

umka2103 [35]

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

$E=\dfrac{kq}{r^2}$

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

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4 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

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What is electrostatic force? ​

What is electrostatic force?