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igor_vitrenko [27]
3 years ago
9

g From the following choices, select the one correct statement: a. Ultraviolet light has a shorter wavelength than infrared ligh

t. b. Blue light has a shorter wavelength than X rays. c. Radio waves have a shorter wavelength than gamma rays. d. Gamma rays have a longer wavelength than infrared waves. e. Blue light has a longer wavelength than red light. 2. The diagram to the left is Fray with no lens in front of her.
Physics
1 answer:
Sati [7]3 years ago
4 0

Answer:

Ultraviolet light has a shorter wavelength than infrared light.

Explanation:

Electromagnetic waves are waves that does nor require material medium for their propagation. Example of Electromagnetic waves are classified according to the acronym RIVUXG. They decrease in wavelength and increase in frequency from Radiowaves to Gamma ray.

R is the Radio waves

U - ultraviolet ray

V - Visible light (ROYGBIV){Red, Orange, Yellow, Green, Blue, Indigo and Violet}

Note that the visible light also decrease in wavelength from red to indigo, therefore red will have longer wavelength compare to blue light.

I - Infrared rays

X - X ray

G - Gamma ray

Based on the above conclusion, it can be inferred that Ultraviolet light has a shorter wavelength than infrared light since Infrared rays comes before the infrared light in the electromagnetic spectrum.

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The magnitude of vector b is 8.58 Unit.

Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.

Applying the Pythagoras theorem

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3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
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Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

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How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
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Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


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