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3 years ago

Please help!!!!

Physics

1 answer:

Dafna11 [192]3 years ago

5 0

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=3\cdot 10^{-9}C$ is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

$E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

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2 years ago

How permeable and porous would an aquifer be?

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3 years ago

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

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3 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

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Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

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$P_2$ = Momentum of the ball after hit

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$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

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$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

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$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

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The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

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3 years ago

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