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antiseptic1488 [7]

3 years ago

14

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

trobe light. what radiation pressure is produced on the cardboard by the light?

1 answer:

PolarNik [594]3 years ago

8 0

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

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People are constantly being barraged with low-level radiation. Which radiation source below contributes the most low-level radia

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3 years ago

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A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0

3 years ago

The dark bands on the wall are from ______________ interference. and i will give brainlyest

Fittoniya [83]

Light can be seen as an electromagnetic wave.
What happens when two waves, with the same frequency, superpose is called interference.

If at a certain point two waves arrive both with a crest, we have constructive interference and the amplitudes sum up, reaching the maximum value, resulting in bright spots.

If at a certain point one of the waves arrives with a crest and the other wave arrives with a trough, we have destructive interference, and the two amplitudes cancel out, resulting in dark spots.

Therefore, t<span>he dark bands on the wall are from destructive interference.</span>

4 0

4 years ago

The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-

WINSTONCH [101]

Answer: $1.212(10)^{-10} m$

Explanation:

The de Broglie wavelength $\lambda$ is given by the following formula:

$\lambda=\frac{h}{p}$ (1)

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$p$ is the momentum of the atom, which is given by:

$p=m_{e}v$ (2)

Where:

$m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg$ is the mass of the electron

$v=6(10)^{6}m/s$ is the velocity of the electron

This means equation (2) can be written as:

$p=(9.11(10)^{-31}kg)(6(10)^{6}m/s)$ (3)

Substituting (3) in (1):

$\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)}$ (4)

Now, we only have to find $\lambda$ :

$\lambda=1.2122(10)^{-10} m$ >>> This is the de Broglie wavelength of the electron

8 0

3 years ago

What would the circuit resistance be if a 7.5 Amp draw was present with the engine running and the charging system producing 15

AlladinOne [14]

Answer:

2 ohms

Explanation:

Hi there!

Ohm's law states that $V=IR$ where V is the voltage, I is the current and R is the resistance.

Plug in the given information (I=7.5, V=15) and solve for R

$V=IR\\15=(7.5)R$

Divide both sides by 7.5 to isolate R

$\frac{15}{7.5}= \frac{7.5R}{7.5} \\2=R$

Therefore, the circuit resistance would be 2 ohms.

I hope this helps!

4 0

3 years ago