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sammy [17]
3 years ago
10

An object is placed in front of a thin converging lens of focal length 12 cm. The image of the object is upright and 2.5 times a

s tall as the actual object. What is the distance of the object to the lens?
7.2 cm
6.0 cm
4.2 cm
15 cm
Physics
2 answers:
DaniilM [7]3 years ago
8 0
Digamos que unos 4.2 centímetros
yuradex [85]3 years ago
7 0

Answer:

The <em><u>distance of the object</u></em> to the lens is

7.2 \: cm

The solution is in the attached image,

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I am stuck on a kinetic energy problem, any help would be greatly appreciated:)))
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When a 4-μf capacitor has a potential drop of 20 v across its plates, how much electric potential energy is stored in this capac
Genrish500 [490]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2.

Substituting numerical value

E=\frac{1}{2}(4*10^{-6}) (20)^2\\ E=800\; \mu J

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4 years ago
You push very hard on a heavy desk, trying to move it. You do work on the desk:
frozen [14]

Answer:

(C) Only if it starts moving

Explanation:

We know that work done is given by

W=F.d=Fdcos\Theta

So there are two case in which work done is zero

First case is that when force and displacement are perpendicular to each other

And other case is that when there is no displacement

So for work to be done there must have displacement, if there is no displacement then there is no work done

So option (c) will be the correct option

3 0
3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
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