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3 years ago

2. The amount of charge and the distance from the charge determine the...

Physics

2 answers:

olasank [31]3 years ago

8 0

Answer:

Strength of the electric field

Explanation:

The amount of charge and the distance from the charge determine the strength of the electric field.

Mathematically, it can be written as :

$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{r^2}$

where,

q is the electric charge

r is the distance between the charges

It can be also defined as the electric force per unit electric charge. Hence, the correct option is (c) " Strength of the electric field ".

cricket20 [7]3 years ago

7 0

Answer:

the answer is c Strength of the electric field

Explanation:

it is that bcause

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As you drive in your car at 22.5 mi/h, you see a child’s ball roll into the street ahead of you. You hit the brakes and stop as

evablogger [386]

Answer:

Explanation:

Speed of car =22.5miles/hr

U=22.5miles/hour

Applied brake and come to rest

Final velocity, =0

t, =2sec

Given that,

Speed=distance /time

Then,

Distance, =speed, ×time

Converting mile/hour to m/s

Given that

Use: 1 mile= 1600 m, 1 h= 3600s

22.5miles/hour × 1600m/mile × 1hour/3600s

Therefore, 22.5mile/hour=10m/s

Using speed =10m/s

Distance =speed ×time

Distance=10×2

Distance, =20m

The distance travelled before coming to rest is 20m.

5 0

3 years ago

The attraction between earth and the moon is an example of ________ force. the attraction between earth and the moon is an examp

Korolek [52]

The answer is gravitational force. The gravitational force between the earth and the moon is the similar as between any other two masses in space.

Newton clarified that the force of attraction between two masses is the outcome of the weight of object one multiplied by the weight of objects two multiplied by the gravitational constant divided by the space between the two masses squared.

4 0

4 years ago

Read 2 more answers

Can toe touches help you improve flexibility? Yes/No

jolli1 [7]

Yes it helps improve flexibility

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3 years ago

Read 2 more answers

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

Circular Motion

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have