<u> Ohms law: </u> This law relates voltage difference between two points. Mathematically, the law states that V=IR;
Where
V = voltage difference ; in volts
I = Current ; in Amperes
R = Resistance ; in ohms
<u>1. Answer : </u> given that R = 10 ; V= 12 V ; I = ?
From ohms law, I = V/R
= 12/10
= 1.2 Amp.
<u>2. Answer:</u> given that R = 10 ; V= ? ; I = 5
From ohms law, V = IR
= 10×5 = 50 V
<u>3 . Answer:</u> given that R = ? ; V= 120 ; I = 5
From ohms law, R = V/I
= 120/5
= 24 Ω
<u>4 . Answer:</u> given that R = ? ; V= 10 ; I = 20
From ohms law, R = V/I
= 10/20
= 0.5 Ω
<u>5 . Answer:</u> given that R = 480 ; V= 24 ; I = ?
From ohms law, I = V/R
= 24/480
= 0.05 A
<u>6. Answer:</u> given that R = 150 ; V= ? ; I = 1
From ohms law, V = IR
= 1 × 150
= 150 V
Answer: The correct statements are:
- The atoms are very attracted to one another.
- The atoms are held tightly together.
Explanation:
Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.
In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.
The key point about solid are:
- The atoms are very attracted to one another.
- The atoms are not moving freely.
- It will not spread out evenly to fill any container.
- The atoms are held tightly together.
- The forces of attraction are strong to bring molecules together.
- The atoms are close and in fixed positions.
Answer:
In physics, work is the amount of energy required to perform a given task (such as moving an object from one point to another). We start by defining the scalar product of two vectors, which is an integral part of the definition of work, and then turn to defining and using the concept of work to solve problems.
Answer:
A stone that is dropped down into an empty well
Answer:

Explanation:
We are given that three resistors R1, R2 and R3 are connected in series.
Let
Potential difference across 
Potential difference across 
Potential difference across 
We know that in series combination
Potential difference ,
Using the formula

Hence, this is required expression for potential difference.