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torisob [31]
2 years ago
14

Give me some pointers please, I need to finish this ASAP-

Physics
1 answer:
Vinil7 [7]2 years ago
3 0
Stop cheating learn from ur resources
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likoan [24]
The answer to this would be ocean floor
4 0
3 years ago
Which of the following is an element?
oksano4ka [1.4K]
Chlorine pls mark as brianliest
7 0
3 years ago
The acceleration due to gravity on earth will decrease as which of the following occurs. The mass of the object decreases. The d
pentagon [3]

Answer:

The distance of the object to the center of the earth increases.

Explanation:

The acceleration due to gravity on Earth is given by:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the object from the Earth's centre

We notice that:

- g does not depend on the mass of the object

- g is inversely proportional to r

This means that if the distance of the object from the Earth's centre increases, g decreases. So, the correct option is

The distance of the object to the center of the earth increases.

4 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
How is a conducter different from an insulater
Maru [420]

Answer:

Conductors have magnetic fields; insulators do not have magnetic fields. Conductors do not have magnetic fields; insulators do have magnetic fields. ... In a conductor, electric current cannot flow freely; in an insulator, it can flow freely.

3 0
3 years ago
Read 2 more answers
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