Why do we use silk threads to suspend charged rods instead of metallic threads?

1 answer:

8 0

Answer:When you rub the glass rod with the silk, the rod charges positive. If you place a charged rod on one of the swivel stands, and then bring another charged

Explanation: i no it

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Two resistors, R1 and R2, are

dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/R₁ + 1/(221.0 Ω)

1/R₁ = 1/(120.7 Ω) - 1/(221.0 Ω)

R₁ = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0

3 years ago

Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce

zysi [14]

Answer:

20 m/s

Explanation:

Given

u(A) = 20 m/s
u(B) = 30 m/s
acceleration equal in magnitude but opposite in direction

Solving

Velocity of A at Q = 30 m/s
From, P to Q, Δv(A) = 30 - 20 = +10 m/s
Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
Δv(B) = v(B at Q) - u(B at P)
-10 m/s = v(B at Q) - 30 m/s
v(B at Q) = 30 - 10 = 20 m/s

6 0

3 years ago

A conductor of radius r, length l and resistivity p has resistance R. It is melted down and formed into a new conductor, also cy

Vaselesa [24]

Answer:

b) R/4 (There seems to an error in mentioning the multiple choices of this question, please see below explanation of correct calculations for this question.)

Explanation:

dimension of the conductor before melting is l, r

reistivity is p

R=(p*l)/(pie*r2)

after reforming length is reduced to L=l/4

volume in both the cases will be same

i.e. pie * r^2 * l =pie * R^2 * L

r^2 * l = R^2 * (1/2)l

due to this radius will become R=sqrt(2) * r

now new reistance is given by Rx=(p * L)/(pie * R^2)

i.e. Rx=(p * l/2)/(pie * r^2 * 2)

after simplification RX=((p * l)/(pie * r^2))/4

i.e. Rx=R/4

5 0

3 years ago

Read 2 more answers

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$