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son4ous [18]
2 years ago
9

1. The potential difference between the terminals a cell in open circuit is 2.2volt. With resistance of 5ohm across the terminal

s of a cell, the terminal potential difference is 1.8 volt. Calculate the internal resistance.​
Physics
1 answer:
nadya68 [22]2 years ago
7 0

Explanation:  

10 /9 Ω

potential difference across the cell in open circuit is the emf of the cell.

Hence, emf E=2.2V

when, circuit is closed, potential difference across cell is given by V=E−Ir

And,

I= E/ R+r

Hence, V= E− Er/ R+r

⟹ V= ER/ R+r

⟹ 1.8= 2.2×5 /5+r

⟹9+1.8r=11

⟹ r= 2/  1.8  Ω

⟹ r= 10/9  Ω

​

​

 

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A brick sits on the top of a hill with a gravitational potential energy of 245 J. To determine the gravitational potential of th
Bogdan [553]

Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

In order to determine the gravitational potential energy of the brick, we must know the mass (m) of the brick, its acceleration due to gravity (g) since it is acting under the influence of gravitational force and the distance between the top of the hill and the ground level. (The height).

Potential energy of a body is calculated as mass × acceleration due to gravity × height.

5 0
3 years ago
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resista
aliya0001 [1]

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

8 0
2 years ago
A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

E = 5.08 x 10^5 N/C

7 0
3 years ago
a 2.5 kg rock is dropped off a 32 m cliff and hits a spring, compressing it 57cm. what is the spring constant
Rama09 [41]

2.5 kg because you cant change the weight of the rock

4 0
3 years ago
Read 2 more answers
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
3 years ago
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