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PolarNik [594]
3 years ago
8

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

of the electric field at this location?
Physics
1 answer:
madam [21]3 years ago
3 0

The magnitude of the electric field is 1.51\cdot 10^{10}N/C

Explanation:

The electrostatic force experienced by an electric charge immersed in an electric field is given by

F=qE

where:

q is the charge of the particle

E is the strength of the electric field

F is the force

In this problem, we have:

q=+4.3\cdot 10^{-18}C is the charge of the particle

F=6.5\cdot 10^{-8} N is the force

Therefore, we can find the electric field by solving the equation for E:

E=\frac{F}{q}=\frac{6.5\cdot 10^{-8}}{4.3\cdot 10^{-18}}=1.51\cdot 10^{10}N/C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

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Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
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Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

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On substituting the known values,

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Final answer:

The parametric equations describing the shot are

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