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2 years ago

In which switching technique, a dedicated

Physics

1 answer:

Aloiza [94]2 years ago

3 0

The answer is Circuit Switching

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Question: the distance of the car can move by velocity 15 m/s for time 5s is : Physics

tangare [24]

Answer:

c) 2.05

Explanation:

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2 years ago

A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free

Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R from equation (1)

T = I x a / R² .... (2)

mg - T = ma .... (3)

Substitute the value of T from equation (2) in equation (3)

$mg - \frac{Ia}{R^{2}}=ma$

$a=\frac{mg}{m+\frac{I}{R^{2}}}$

T is the acceleration in the system

Substitute the value of a in equation (2)

$T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}$

$T=\frac{I\times mg}{I+mR^{2}}$

This is the tension in the string.

4 0

2 years ago

Dont know what to type here ok chat meh

never [62]

ogckcigxxkgxkgxkgxixogxigxixgxgifkcgo

5 0

2 years ago

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with

Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

The net charge enclosed for each r in this range is positive and the electric field is outward

For r>2R

Charge enclosed is zero, so electric field is zero

8 0

2 years ago

In the diagram, which is demonstrated by arrow 1? A. liquid water evaporating from the surface to the atmosphere B. water vapor

Shkiper50 [21]

I think D. liquid water moving along the surfac

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2 years ago