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3 years ago

5

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

lue of the magnetic field in the wave

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

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Five metal samples, with equal masses, are heated to 200oC. Each solid is dropped into a beaker containing 200 ml 15oC water. Wh

Ksju [112]

Part 1) Which metal will cool the fastest?
To answer this question, we should have a look at the formula of the heat flow rate, which says "how fast" a material is able to heat/cool:
$\frac{\Delta Q}{\Delta t} = -k \frac{A \Delta T}{x}$
where:
$\Delta Q$ is the heat exchanged
$\Delta t$ is the time interval
k is thermal conductivity of the material
A the surface where the exchange of heat occurs
$\Delta T$ the variation of temperature
x is the thickness of the material
We see that the heat flow rate $\frac{\Delta Q}{\Delta t}$ is linearly proportional to k, the thermal conductivity of the material. So, the larger k, the fastest the metal will cool.
If we have a look at the thermal conductivity of each metal, we find:
- Aluminium: 237 W/(mK)
- Copper: 401 W/(mK)
- Gold: 314 W/(mK)
- Platinum: 69 W/(mK)
Therefore, copper is the material with highest heat flow rate, so the metal which cools fastest.

Part 2) Which sample of copper demonstrates the greatest increase in temperature
To solve this part, we can have a look at how the amount of heat exchanged Q is related to the increase in temperature $\Delta T$ :
$Q=m C_S \Delta T$
where m is the mass and Cs the specific heat of the material. Re-arranging the formula, we get
$\Delta T= \frac{Q}{m C_s}$
therefore, we see that the increase in temperature is inversely proportional to the mass m. This means that the block that will show the largest increase in temperature is the block with the smallest mass, so the correct answer is A) 0.5 kg.

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4 years ago

A bicycle has wheels of 0.70 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 10.8 s. Wh

Anna11 [10]

Answer:

$\alpha=0.16\frac{rad}{s^2}$

Explanation:

Angular acceleration is defined as the variation of angular speed with respect to time:

$\alpha=\frac{\omega}{t}$

The relation between the angular speed and the linear speed is given by:

$\omega=\frac{v}{r}$

Replacing (2) in (1):

$\alpha=\frac{v}{rt}$

We need to convert $\frac{km}{h}$ to $\frac{m}{s}$ :

$22\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=6.11\frac{m}{s}$

Recall that:

$r=\frac{d}{2}=\frac{0.7m}{2}=3.5m$

Replacing:

$\alpha=\frac{6.11\frac{m}{s}}{(3,5m)(10.8s)}\\\alpha=0.16\frac{rad}{s^2}$

3 0

4 years ago

Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying

Tanya [424]

10,378.82 m/s is the second cosmic speed.

69,801 km is the radius of the synchronous orbit of a satellite.

Given

Mass of planet = 4.74 × $10^{24}$ kg

Radius of planet = 5870 km = 5870000m

First Cosmic speed = 7.34 km/sec

1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.

It can be calculated by

v = √2GM/r where,

v= Escape velocity of the satellite

G = Gravitational constant

M = Mass of planet

r = Radius of planet

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

2) Speed of the satellite at the given period

v = 2πr/T where,

T= Time period of rotation = 16.6 × 3600 seconds

r = v×T/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Hence

The <u>Second Cosmic Speed</u> i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.

And the radius of the synchronous orbit of a satellite is 69,801 km.

Learn more about cosmic speed here brainly.com/question/15351190

#SPJ1

3 0

2 years ago

A car traveling at 22m/s is in an accident, and the 81 kg driver is brought to a stop in 1.4 s by in inflating airbag. What forc

larisa86 [58]

Here, F = m * a
F = m * v/t
Here, m = 81 Kg
v = 22 m/s
t = 1,4 s

Substitute their values,
F = 81 * 22/1.4
F = 81 * 15.71
F = 1273 N
So, Closest value from your options is 1300 N

In short, Your Answer would be Option B

Hope this helps!

8 0

4 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago