I have all the answers here so take this
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B

Where
=Mass of block B
Substitute the values


For block A

Where
Mass of block A
Substitute the values


Hence, the friction force acting on block A=40 N
Answer:
Explanation:
Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree
From the formula of resultant vector
(1.703I)² = I² + I² + 2 I² cosФ
2.9 I² = 2I² + 2 I² cosФ
.9I² = 2 I² cosФ
cosФ = .9 / 2
= .45
Ф = 63.25 .
Answer:
Explanation:
Distance between plates d = 2 x 10⁻³m
Potential diff applied = 5 x 10³ V
Electric field = Potential diff applied / d
= 5 x 10³ / 2 x 10⁻³
= 2.5 x 10⁶ V/m
This is less than breakdown strength for air 3.0×10⁶ V/m
b ) Let the plates be at a separation of d .so
5 x 10³ / d = 3.0×10⁶ ( break down voltage )
d = 5 x 10³ / 3.0×10⁶
= 1.67 x 10⁻³ m
= 1.67 mm.
The answer is that FRICTION PRODUCES HEAT.