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3 years ago

What factor is responsible for altering the speed of an electromagnetic wave?

Physics

2 answers:

cupoosta [38]3 years ago

4 0

Answer: The correct answer is "type of medium".

Explanation:

Electromagnetic wave is the wave in which both electric field and the magnetic field are mutually perpendicular each other to the motion of the wave. It travels with the speed of light.

When the electromagnetic wave travels from one medium to the another medium then the speed of the wave will get change and therefore, it changes the wavelength.

Therefore, the type of medium is responsible for altering the speed of an electromagnetic wave.

irina1246 [14]3 years ago

3 0

The answer is c. types of medium

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As a ball bounces from a floor, its acceleration off the floor between bounces is

frez [133]

Its acceleration off the floor keeps reducing/diminishing

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3 years ago

Two identical, flat, circular coils of wire each have 100 turns and a radius of 0.500 m. The coils are arranged as a set of Helm

kvv77 [185]

Answer:

The magnitude of the magnetic field at the point of interest is B=1.8·10^-3T

Explanation:

To solve this problem we can calculate the expression of the magnetic field for one of the coils and using the symmetry of the problem we can use twice this expression to obtain the expression of the total field as:

$B_{T}=B_{1}+B_{2}$

For the magnetic field $B{1}$ we use the Biot-Savart law:

$d\vec{B}=\frac{\mu_{0}i\vec{dl}}{4\pi}\times \frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^3}$

Using cylindrical coordinates, the expression of the magnetic field in a circular coil is:

$B_{i}=\frac{\mu_{0}ir^2}{2(\sqrt{z^2+r^2})^3}$

The expression of the magnetic field in a circular coil with N turns is:

$B_{i}=\frac{\mu_{0}iNr^2}{2(\sqrt{z^2+r^2})^3}$

Because we have a point of symmetry in the middle between the 2 coils, the expression of the field in that point is:

$B_{T}=B_{1}+B_{2}=\frac{\mu_{0}iNr^2}{(\sqrt{z^2+r^2})^3}$

with r=0.5m, Z=0.25m, i=10A and N=100

4 0

3 years ago

Why can't we breath under water?

Kipish [7]

We can certainly draw water into our lungs, even though
our brain screams "Don't do that !".

But our lungs can only separate oxygen out of air, not out of
water. So if there's only water in our lungs, we pass out after
a short time because we're not getting any oxygen.

5 0

3 years ago

Read 2 more answers

What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of v and g. Use three significant fig

max2010maxim [7]

The speed u of the object at the height of (1/2) hmax is $0.707$ $v_{i}$ .

<u>Explanation:</u>

At a height that is half of the maximum height, the object will have only half of its initial kinetic energy, the rest half is converted to potential energy at this point. So,

$1 / 2 m v^{2} _{i} = m (v_{mid})^{2} }$

$v_{mid} = \frac{v}{\sqrt{2} }$

$v_{mid} =$ $0.707$ $v_{i}$ .

The speed u of the object at the height of (1/2) hmax is $0.707$ $v_{i}$ .

3 0

3 years ago

A certain rocket of initial mass mo carries 60% of its initial mass as fuel. What is its final speed if it accelerates from rest

alexgriva [62]

Answer:

A) $\frac{\Delta v}{ve} = 0.91$

B) $\frac{\Delta v}{ve} = 1.05$

Explanation:

Tsiolkovsky rocket equation states that:

$\Delta v = ve * ln(\frac{m0}{mf} )$

Where

Δv: change of speed of the rocket after burning the fuel

ve: exhaust velocity

m0: initial mass of the rocket

mf: mass of the rorcket after completing the burn (if it burned all its fuel this is the dry mass)

Since the rocket starts from rest (v = 0) its final speed will be the Δv

To express this speed in multiples of the fuel velocity out of the rocket, we simply:

$\frac{\Delta v}{ve} = ln(\frac{m0}{mf} )$

In this case we know that 60% of the mass of the rocket is fuel, so:

$mf = m0 * (1 - 0.6) = 0.4*m0$

So:

$\frac{\Delta v}{ve} = ln(\frac{m0}{0.4*m0} ) = ln(\frac{1}{0.4} ) = 0.91$

If the rocket jettisons a first stage with 0.3*m0 of fuel, and 0.1*m0 of fuel tank, it will be doing two burns.

The first burn will go from mass m0 to m1

m1 = m0 - 0.3 * m0 = 0.7 * m0

The speed gain from this first burn would be:

$\frac{\Delta v1}{ve} = ln(\frac{m0}{m1} )$

Then it will jettison the tank, since jettisoning is done with small explosives the gain of speed can be ignored for being very small. However it would have lost some mass.

m2 = m1 - 0.1 * m0 = 0.7 * m0 - 0.1 * m0 = 0.6 * m0

Next it will activate the second stage, burning fuel and going from m2 to m3

m3 = m2 - 0.3 * m0 = 0.6 * m0 - 0.3 * m0 = 0.3 * m0

The speed gain of this second burn would be:

$\frac{\Delta v2}{ve} = ln(\frac{m2}{m3} )$

And both added together:

$\frac{\Delta v}{ve} = ln(\frac{m0}{m1} )+ ln(\frac{m2}{m3} )$

$\frac{\Delta v}{ve} = ln(\frac{m0}{0.7 * m0} )+ ln(\frac{0.6 * m0}{0.3 * m0} ) = 0.356 + 0.693 = 1.05$

3 0

3 years ago