B. Starting with most recent
Answer:
Change in mechanical energy = work done by friction
so it is equal to

Explanation:
As we know that change in mechanical energy must be equal to the work done by non conservative forces only
So here when block moves down the inclined plane then the work done by friction force is given as

here we have

here we know that

so we have


Now the friction force on the block is given as



now work done by the friction is given as


Answer
given,
Echo time = t = 1000 s
speed = ?

a) speed of electromagnetic wave

wave travels Venus two time



d = 1.5 x 10¹¹ m
b) now, echo time



t = 5 x 10⁻⁷ s
Answer:
i. 6.923 V
ii. The e.m.f. = 22.5 V
Explanation:
i. The given parameters are;
Length of potentiometer = 1 m
The resistance of the potentiometer = 10 Ω
The e. m. f. of the attached cell = 9 V
The current, I flowing in the circuit = e. m. f/(Total resistance)
The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A
The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire
The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V
ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

Where:
E = e.m.f. of the balance point cell
= Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω
= Resistance of the cell in the circuit = 3 Ω
V = e.m.f. attached cell = 9 V

E = 7.5*3 = 22.5 V
The e.m.f. = 22.5 V
Answer:
Push because the force you use up is greater then down