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3 years ago

If a negative charge is released in a uniform electric field move

Physics

1 answer:

Goryan [66]3 years ago

8 0

Answer:

Explanation:

Let the electric field strength is E and the charge is q. The force experienced by the charge due to the presence of electric field is

F = q x E

The direction of force is same as that of the electric field strength if the charge is positive in nature and if the charge is negative in nature the direction of force is opposite to the direction of electric field.

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During a storm, the power went out at Bailey’s house. She had no electricity. What could she use as an alternate form of electri

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C a battery-operated lamp sorry if incorrect

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4 years ago

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4 years ago

Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.

Lelechka [254]

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ $T_1Cos30^{\circ}-T_2Cos60^{\circ}=0$

or,

$T_1Cos 30^{\circ}=T_2Cos60^{\circ}$

On substituting the values, we get

$T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}$

$T_1\times \sqrt{3} =T_2$ ....(equation 1)

On the Y-axis,

⇒ $T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N$

$\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N$

$T_1+\sqrt{3}T_2=139.2\times 2$

From equation 1, we get

$T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N$

$T_1+3T_1=278.4 \ N$

$4T_1=278.4 \ N$

$T_1=\frac{278.4}{4}$

$=69.6 \ N$

6 0

3 years ago

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230

slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

$V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}$

Substitute the given values

$q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC$

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

$=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2$

σ=17 nC/m²

3 0

3 years ago

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

Elina [12.6K]

Before going to answer this question first we have to know the fundamental principle of magnetism.

A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.

Through this concept the question can be answered as explained below-

A-As per first option the side of magnet A is repelled by the south pole of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.

B-The side of magnet A is repelled by the north pole of magnet B. Hence the side of A must be north pole.It can't be a south pole.

C-The side of magnet A is attracted by the south pole of magnet B .Hence the side of magnet A must be north.Hence this is right

D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.

Hence the option C is right.

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3 years ago

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