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3 years ago

The wavelength corresponding to light with a frequency of 4 x 10^14 Hz is 1.33m 0.075 mm 7500 nm 750 nm

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

Wavelength of light is 750 nm.

Explanation:

It is given that,

Frequency of light, $\nu=4\times 10^{14}\ Hz$

The relationship between the wavelength and the frequency of light is given by :

$c=\nu\times \lambda$

Where

c = speed of light

$\nu$ = frequency of light

$\lambda$ = wavelength of light

$\lambda=\dfrac{c}{\nu}$

$\lambda=\dfrac{3\times 10^8\ m/s}{4\times 10^{14}\ Hz}$

$\lambda=7.5\times 10^{-7}\ m$

$\lambda=750\ nm$

Hence, the correct option is (d) "750 nm".

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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

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a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

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3 years ago

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Answer:

a) 3.56 x 10^22 N

b) 3.56 x 10^22 N

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Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = $\frac{GMm}{R^2}$

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = $\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}$ = 3.56 x 10^22 N

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N

b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N

7 0

3 years ago