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ankoles [38]
3 years ago
6

To understand the electric potential and electric field of a point charge in three dimensions Consider a positive point charge q

, located at the origin of three-dimensional space. Throughout this problem, use k in place of 14?
Physics
1 answer:
Liula [17]3 years ago
5 0

Answer:

The magnitude of the electric field at distance r from the point charge q is \dfrac{kq}{r^2}

Explanation:

Suppose, Find E(r), the magnitude of the electric field at distance r from the point charge q.

We know that,

The electric field is proportional to the point charge and inversely proportional to the square of the distance from the charge point .

In mathematically,

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

Where, q = charge point

R = distance from charge point

Hence, The magnitude of the electric field at distance r from the point charge q is \dfrac{kq}{r^2}

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nignag [31]

Answer:

How fast is it going? 29.4 Meters per second

How far has it fallen? 44.1 meters

Explanation: Gravitional Acceleration: 9.8 meters per secnd squared!

6 0
3 years ago
needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene
Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

  <em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ? 

Compared to gravity on Earth, it's only  16.5 percent as much on the Moon. 
So naturally, from the formula, you'd expect the Potential Energy to be less
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4 0
3 years ago
A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s
Jlenok [28]

Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

4 0
3 years ago
I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation
ludmilkaskok [199]

Answer:

-5.39\times10^{33} J

Explanation:

Potential energy =

-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J

7 0
2 years ago
Which one of the following statements does not accurately describe vibrations?
Ivanshal [37]

This is an example of resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The result of resonance is always a large vibration.

Answer D. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body Because this statement contridicts the above statement, it is not accurate

6 0
4 years ago
Read 2 more answers
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