3 years ago

Which of the following properties could be described as a force that pushes charges along?

Physics

1 answer:

Harlamova29_29 [7]3 years ago

7 0

C) voltage is a force that pushes charge along

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The combination of all forces that act on an object are (2 points)

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Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

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4 years ago

The water portion of Earth; one of the traditional subdivisions of Earth’s physical environment

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hydrosphere

hope this helps :)

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3 years ago

Read the following experiment. Describe the experiment's results; identify which of Newton's laws is illustrated and how these r

Volgvan

It’s 3 guaranteed because if you flick the card horizontal it’s going to bend the laws

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3 years ago

If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0

sineoko [7]

Answer:

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

Explanation:

u = Object distance = 9 m

v = Image distance = 1.7 cm (as the image is forming on the retina)

$h_u$ = Object height = 1.5 m

Magnification

$m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m$

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

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3 years ago

A small particle has charge -2.00 uC and mass 1.50×10-4 kg. It moves from point A, where the electric potential is V(A) = 200 V,

Svet_ta [14]

Answer:

$v_b=6.13 m/s.$ .

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

$T.E=qV_a+\dfrac{1}{2}mv_a^2=qV_b+\dfrac{1}{2}mv_b^2\\\dfrac{1}{2}mv_b^2=\dfrac{1}{2}mv_a^2+q(V_a-V_b)\\$

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get, $v_b=6.13 m/s.$ .

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

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4 years ago