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podryga [215]
3 years ago
6

Exerts a force on anything that has an electric charge

Physics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

Electric Field

Explanation:

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Help please (20 pts) I think it's C
gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

6 0
3 years ago
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: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

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
. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe
Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0
3 years ago
What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?
Elodia [21]

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

8 0
2 years ago
Read 2 more answers
An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement
dimaraw [331]

Answer and Explanation:

distance will be 2×3.14 (pie)×r

displacement will be 2r (diameter)

the motion is uniform circular motion as the object is moving in a circular path with uniform motion

8 0
2 years ago
Read 2 more answers
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