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3 years ago

Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.

Physics

2 answers:

VladimirAG [237]3 years ago

8 0

Answer:

Unlike Charges:

The unlike charges attracts each other, as we can say that the two atoms or point charges which posses different properties attracts one another.

Explanation:

The electric field lines:

The electric fields shows the direction, the nature, and the magnitude of the charges been placed there in a given space. As the field lines from the positive point charge(+) goes in the outward direction and the negative charges(-) has the field lines coming towards it or they are directed towards inside the test charge.

As, the different charges interact lines when interact with one another, the lines become more curved and they slide over the two charges. Now, that is also due to the number of charges attracted toward each present on the surface of the two charges.

harina [27]3 years ago

4 0

Hello.

The answer is:

It creates a spark.

Then thespark can sometimes start a fire or other serious problems.

Have a nice day

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4 years ago

6a. A special lamp can produce UV radiation. Which two statements

alexandr1967 [171]

Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

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4 years ago

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p

Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m$

The image distance is 0.051 m

When u = 50 cm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m$

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

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3 years ago