Answer:
I for a rod about its center is I = M L^2 / 12
A = L (1/2 - 1/3) = L / 6 distance from middle to location L/3
A^2 = L^2 / 36
I = M L^2 (1/12 + 1/36) = M L^2 (4 / 36) = M L^2 / 9
I about given location by parallel axis theorem
Answer:
Explanation:
We shall apply conservation of momentum along x and y axis.
Let the final momentum of second particle be p₁ along x axis and p₂ along y axis.
Considering momentum along x axis
2 + 0 = 3 cos 45 + p₁
p₁ = 2-2.12 = - 0.12 kg m/s
Considering momentum along y axis
4 + 0 = 3 sin 45 + p₂
p₂ = 4-2.12 = 1.88 kg m/s
Final momentum = √ ( p₁² + p₂² )
=√ ( .12² + 1.88² )
= 1.88 approx
Answer:
Explanation:
The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark.
If the resistor in the circuit is 1 megaohlm, the total resistance in daylight and darkness will be 1.01 megaohms and 1.1 megaohlms.
The percentage difference = (1.1-1.01)/1.1*100% = 8.18%
If the resistor in the circuit is 25 kohlm, the total resistance in daylight and darkness will be 35 kohms and 125 kohlms.
The percentage difference = (125-35)/125*100% = 72%
With the input p.d to the sensing circuit fixed at 12 v, the sensing current will change according to the total resistance. A 72% difference is much more detectable. So the 25 kohm resistor is the better choice.
Answer:i
9E13 ELECTRONS
Explanation:
First we find the number of charges in two hrs
Which is - 2x 7200=- 14400NC
So no of electrons is now q/e
-144x10^-7/-1.6*10^-19
= 9*10^13