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-Dominant- [34]
2 years ago
13

Impact of electricity in the society

Physics
1 answer:
Travka [436]2 years ago
5 0

Answer:

<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>

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The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat
Nonamiya [84]

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

7 0
3 years ago
Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain
algol13

Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

The initial mass of the sample = 1 microgram

After 5 half-lives, the mass should reduce to 1/32 of the original.

So the concentration left = 1/32 of 1 microgram = 1/32 microgram

7 0
3 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
2 years ago
When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart
Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

I=\Delta p=m\Delta v

where

m = 5.0 kg is the mass of the cart

\Delta v=2.2 m/s is its change in speed

Substituting numbers into the equation, we find

I=(5.0kg)(2.2 m/s)=11 kg m/s

7 0
3 years ago
Read 2 more answers
MathPhys i need your help please helpppo
Lena [83]

Answer:

6.77 m/s

Explanation:

First, in the x direction:

Given:

Δx = 3.17 m

v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

Δx = v₀ t + ½ at²

(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²

3.17 = 0.859 v t

3.69 = v t

Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables.  Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

66.8 / v² = 1.458

v² = 45.8

v = 6.77

7 0
3 years ago
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