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3 years ago

15

A negative test charge will accelerate toward regions of electric potential and electric potential energy.

1 answer:

dalvyx [7]3 years ago

5 0

Answer: higher and lower

Explanation:

charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential the direction of decreasing electric potential energy is the direction of increasing electric potential.

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Question 2 of 34

tatiyna

You can find the magnitude of the resultant vector : (B). By adding the magnitudes of the two vectors

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A vector can be defined as any quantity which possesses magnitude and also has direction.

A Vector quantity is very useful because This type of quantity gives more details to the student or teacher analyzing it.

Lear more about Vectors: brainly.com/question/25705666

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2 years ago

Which describes electricity?

SVETLANKA909090 [29]

Answer:

all of the above

Explanation:

- a build up of electric charge

- the force and motion of electrically charged particles

- an electric current

are three different ways to describe electricity.

So the answer is all of the above.

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2 years ago

What forms when two of earths plates pull apart

zhannawk [14.2K]

Answer: (Screen Shot)

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3 years ago

Read 2 more answers

A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

AlekseyPX

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

3 0

3 years ago

A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang

satela [25.4K]

Nah a a yuloure

Explanation:

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3 years ago