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2 years ago

5. Is it likely that light with frequencies higher than ultraviolet was

Physics

1 answer:

Vinvika [58]2 years ago

3 0

Answer: YES

Explanation:

If Tia Ana exposes her eyes to light ray in which it frequency is higher than ultraviolet ray, it may result to a gradual eyes issue such as cataracts as the radiation damages the cornea and the lens in front of the eyes.

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True or False: Chemical energy stored in food cannot be transformed into mechanical energy

Novosadov [1.4K]

The answer to this question is false.

7 0

2 years ago

Read 2 more answers

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

2 years ago

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 kg of refrigerant flo

kozerog [31]

Answer:

5.15J

Explanation:

First. 54% of the 7kg refrigerant is liquid

So we find mass of vapour at inlet generator

M1 = ( 1-0.54)*7= 3.2kg

At compressor mass of vapour will be

M2= 0.95*7= 6.7kg

So the Mass of vapour at exit generator is

M2-M1= 3.5kg

So to find heat absorbed by refrigerant in evaporation

Its using

Q= mh

°= 3.5x 1.50×10^5 J/kg

=5.15J

4 0

3 years ago

Where is the switch located on this diagram?

Fed [463]

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

6 0

2 years ago

an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object

emmainna [20.7K]

Answer:

The acceleration of the object is $-30\ m/s^2$