Explanation: 7.7 is the distance and an hour and a half is the average speed. You would have to divide the total distance by the total time. So it would be 7.7 divided by 1.5 which would equal 5.13333333333...

6 0

4 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago

An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.

soldier1979 [14.2K]

<u>The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s</u>

Data given;

Mass = 4kg

Initial velocity = 20m/s

length CD = 5m (from the image given)

<h3>Determine The Length of YD</h3>

$YD=CDsin$ θ = $5sin45=\frac{5}{\sqrt{2} } = 3.57m$

<h3>The velocity of the object at point D</h3>

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2m $v^2$ = mgh $_1$ - mgh $_2$ + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 m $v^2$

$\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s$

The velocity of the object at D with a distance of 5m.

<h3>The the required for the object to reach ground</h3>

The velocity of the object in the y-axis is

$v_y=vsin45=19.09$

Acceleration in y-axis = 9.8

Height = 3.57m

h = $ut+\frac{1}{2}at^2$

$3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}$

Taking the positive value

$\frac{-19.28+21.02}{9.8}=0.17s$

The time required for the object to reach ground is 0.17s

From the calculations above,<u> the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s</u>

Learn more on projectile motion here;

brainly.com/question/1130127

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5 0

3 years ago

What is the difference between a wave and a particle?.

Nimfa-mama [501]

Answer:

See below.

Explanation:

The difference between these two is that a wave is a disturbance of some quantity in space or intermediate while a particle has a definite mass concentrated on a small area.

6 0

2 years ago

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