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GREYUIT [131]
3 years ago
10

Four particles are positioned on the rim of a circle. The charges on the particles are 10.500 mC, 11.50 mC, 21.00 mC, and 20.500

mC. If the electric potential at the center of the circle due to the 10.500 mC charge alone is 4.50 3 104 V, what is the total electric potential
Physics
1 answer:
gregori [183]3 years ago
7 0

The total electric potential will be 4.50 ×10⁴ V. The potential difference is the difference in electric potential between two charged substances.

<h3>What is the electrical potential difference?</h3>

The amount of work performed in an electrical field to move a unit charge from one location to another is defined as the electrical potential difference.

The distance between all charges and the center is the same. The 10,500 mC, 11.50 mC, 21.00 mC, and 20,500 mC charges neglect each other's potentials.

Hence, the total electric potential will be 4.50 ×10⁴ V.

To learn more about the electric potential difference, refer to the link;

brainly.com/question/9383604

#SPJ1



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6 0
3 years ago
Please i really need help! :/
valina [46]
Answer: D

Explanation:
Let us examine the given actions to see which ones generate heat and sound energy from mechanical energy.

A) Stretching a string.
The mechanical stretching creates tension in the string, which is released when the tension is removed. The generation of thermal or sound energy is minimal or negligible.

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The sponge ball experiences compressive loading. This generates minimal or no heat and sound energy.

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This is the correct situation.


8 0
3 years ago
Read 2 more answers
A bolt is to be tightened with a torque of 7.0 N · m. If you have a wrench that is 0.55 m long, what is the least amount of forc
kodGreya [7K]

To solve this problem we will apply the given concept for torque which explains the relationship between the force applied and the distance to a given point. Mathematically this relationship is given as

\tau = F*d \rightarrow F = \frac{\tau}{d}

Where,

\tau = Torque

F = Force

d = Distance

Our values are given as,

\tau = 7.0Nm , d = 0.55m

Therefore replacing we have that the force is

F = \frac{7}{0.55}

F = 12.72N

Therefore the least amount of force that you must exert is 12.72N

4 0
3 years ago
Find the magnitude of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it
laila [671]

Answer: F = 2.1 x 10^-4N

Explanation: Question is incomplete.

The complete question is; A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east.

Given parameters; l = 2.5m, I = 1.5A, B = 0.55 guass = 0.55 x 10^-4 Tesla , theta = 90 (from West to East), F = ?

F = BILsin(theta)

F = 0.55 x 10^-4 x 1.5 x 2.5 x sin 90

F = 2.1 x 10^-4 N.

According to right hand rule, it's direction is upward.

6 0
4 years ago
PLEASE HELP! 7TH GRADE SCIENCE!
professor190 [17]
They travel at different speeds
6 0
3 years ago
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