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2 years ago

If the radiant energy from the Sun comes in as a plane EM wave of intensity 1340 W/m2, calculate the peak values of E and B

Physics

1 answer:

Nastasia [14]2 years ago

5 0

Answer:

Given that

I = 1340 W/m²

µo = 4πx 10^-7 Tm/A

sc = 3 x 10^8 m/s

So to find the peak values of B we use

I = (c Bm²) / (2 µo)

1340 = (3 x 10^8 x Bm²) / (2 x 4πx 10^-7)

1340 = (3 x 10^8 x Bm²) / (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 1340 x (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 3.366 x 10^-3

Bm² = 3.366 x 10^-3 / 3 x 10^8

Bm² = 1.122 x 10^-11

Bm = 3.4 x 10^-6 T

Also to find the peak values of E we

use Em = c x Bm

Em = (3 x 10^8) (3.4 x 10^-6)

Em = 1.005 N/C

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According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai

GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

$\Delta x=1.64\times 10^{-8}\ m$

So, the uncertainty in its position is $1.64\times 10^{-8}\ m$ . Hence, this is the required solution.

6 0

3 years ago

In which of the following conditions would there be the most resistance?

Natasha2012 [34]

Answer:

A) A warm wire

Explanation:

A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.

8 0

2 years ago

A large fake cookie sliding on a horizontal surface isattached to one end of a horizontal spring with spring constantk = 425 N/m

omeli [17]

Answer:

Explanation:

spring constant k = 425 N/m

a ) At the point of equilibrium

restoring force = frictional force

= kx = 10 N

425 x = 10

x = 2.35 cm

b )

Work done by frictional force

= -10 x 2.35 x 10⁻² x 2 J ( Distance is twice of 2.35 cm )

= - 0.47 J

= Kinetic energy remaining with the cookie as it slides back through the position where the spring is unstretched .

= 425 - 0.47

= 424.53 J

4 0

3 years ago

A crew on a spacecraft watches a movie that is two hours long. The spacecraft is moving at high speed through space. An Earth-ba

laiz [17]

Answer:

The duration of the movie is longer than 2 hrs.

Explanation:

Given:

The duration of the movie observed by the crew on the spacecraft is 2 hrs.

According to time-dilation formula:

$t = \dfrac{t_{0}}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}$

Here, $t$ is the required time, $t_{0}$ is the original time, $v$ is the velocity of the spacecraft and $c$ is the velocity of light.

Since $v$ , so $t_{0} > t$ .

So the time required will be large.

4 0

3 years ago

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

Burka [1]

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = $9.232 g/cm^3$

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 3.35\times 10^{-8} cm$

Atomic radius of the polonium in unit cell = r

r = 0.5a

$r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm$

$1 cm = 10^{10} pm$

$1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm$

Approximate atomic radius for polonium-209 is 167.5 pm.

7 0

3 years ago