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Artemon [7]
3 years ago
11

How does the electrical force relate to the charge of an object?

Physics
2 answers:
shtirl [24]3 years ago
6 0

Answer:

It is directly proportional to the charge

Explanation:

The magnitude of the electrical force on a charge is given by:

F=qE

where

q is the magnitude of the charge

E is the magnitude of the electric field acting on the charge

From the formula, we see that F (the force) is directly proportional to q /the magnitude of the charge), so the correct answer is

It is directly proportional to the charge

We can also add something about the direction of the force:

- when the charge is positive, the force has positive sign, which means that its direction is the same as the direction of the electric field (repulsive force)

- when the charge is negative, the force has negative sign, which means that its direction is opposite to the direction of the electric field (attractive force)

mezya [45]3 years ago
4 0

<span>The electric force between two object is directly proportional to the net charge on each object and inversely proportional to the square of the distance between them. <span>So, the answer is: The electrical force is directly proportional to the charge.</span></span>

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A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is
puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}

\dfrac{1}{u}=\dfrac{103}{4080}

u =\dfrac{4080}{103}

The magnification is

m = \dfrac{-v}{u}

m=\dfrac{-240\times103}{4080}

m = -6.05

Hence, The magnification is -6.05.

6 0
2 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.
jolli1 [7]
~Hello there! ^_^

Your question: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.

Your answer: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for violet rays.


Hope this helps~

5 0
2 years ago
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elixir [45]

Answer:

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Explanation:

3 0
2 years ago
Is visible light a form of<br> radiation? Explain.
natali 33 [55]

Answer:

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Explanation:

6 0
2 years ago
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