Electromagnetic radiation travels as

What would be new resistance if length of conductor is doubled and thickness is halved <br>

nignag [31]

Answer:

The new resistance comes out to be = 4 times of original resistance .

6 0

3 years ago

Consider a particle with unit charge q, and mass m, in a constant magnetic field B directed along the positive z–axis. The parti

max2010maxim [7]

Answer:

it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

Explanation:

Magnetic field is along z axis while velocity is in x-z plane

so we will have

$F = q(\vec v \times \vec B)$

so here we can say

$F = q(u\hat i + w\hat k) \times (B \hat k)$

so we will have

$F = quB(-\hat j)$

so here the net force on the charge is perpendicular to its x directional velocity along - Y direction

So due to this component of motion it will move along a circle while other component of the motion will remain uniform always

So here it is combination of two parts

1) Uniform circular motion

2) Uniform motion

So we can say that it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

4 0

3 years ago

A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro

german

Answer:

20m

6.9s

Explanation:

The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.

$E_k = E_p$

$0.5mv^2 = mgh$

$h = \frac{0.5v^2}{g}$

we can subtitute v = 20m/s and g = 10m/s2

$h = \frac{0.5*20^2}{10} = 20 m$

So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.

The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

t = v / g = 20 / 10 = 2s

Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff

$d = \frac{gt_2^2}{2}$