What is the application for reflection of waves?

Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the

Ugo [173]

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

5 0

2 years ago

The formula for speed is v = s/t. <br> a. True<br> b. False

crimeas [40]

If you mean S is the distance then it is true
Velocity = Distance / time

3 0

2 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

You know what movie is rly good the great gastby and also ford v Ferrari

jolli1 [7]

Answer:HETJHETHRTHR

Explanation:

6 0

2 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

2 years ago