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KIM [24]
3 years ago
10

What happens to the charge on the conductive sphere when it is connected to a source of charge such as the electrostatic voltage

source?
Physics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer and Explanation:

The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.

Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.

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When a bar magnet is broken into two pieces each piece has a north and south pole because
Serggg [28]

Answer:

b

every magnet is made of many aligned smaller magnets

Explanation:

6 0
3 years ago
A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i
Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0
3 years ago
Determine the amount of potential energy of a 5N book that is 1.5m high on a shelf.
Alex73 [517]

Answer:

Potential energy of book = 7.5 J

Explanation:

Given:

Weight of book = 5 N

Height of shelf = 1.5 meter

Find:

Potential energy of book

Computation:

Weight = Mass x Acceleration of gravity

Mass x Acceleration of gravity = 5 N

Potential energy = Mass x Acceleration of gravity x Height

Potential energy of book = Mass x Acceleration of gravity x Height

We know that;

Mass x Acceleration of gravity = 5 N

So,

Potential energy of book = 5 x 1.5

Potential energy of book = 7.5 J

3 0
3 years ago
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
4 WILL GIVE YOU BRAINLIST
Arisa [49]
The last one, handmade gifts require more of the givers time!
5 0
3 years ago
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