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Rufina [12.5K]
3 years ago
13

Which of the following statements is true? A. X-rays have longer wavelengths than microwaves. B. Radio waves have shorter wavele

ngths that X-rays. C. Gamma rays have longer wavelengths that UV rays. D. Gamma rays have shorter wavelengths than microwaves.
Physics
1 answer:
3241004551 [841]3 years ago
3 0
D is correct, Gamma rays have the shortest wavelength.
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What is the energy of a single photon of visible light with a wavelength of 500.0 nm? 1 nm=10^−9 m
Dmitrij [34]

Answer:

3.97×10^−19 J

From the planks equation

E=hv

V= c/ wave length

V= 3×10^8/500×10^-9

=6×10^14

E= hv

6.63×10^-34×6×10^14

= 3.97×10^−19 J

Explanation:

8 0
2 years ago
A restaurant records the number of tables served each night, and the results have the values: minimum = 3, lower quartile = 14,
dybincka [34]

To choose the correct box plot, verify each of the options and make sure all the values in the plot match the values provided.

<h3>How to identify the median?</h3>

In a box plot, this value is represented by a vertical line located in the middle of the graph.

<h3>How to identify the maximum and the minimum?</h3>

The maximum is the value located on the farthest right, while the minimum is located on the farthest left.

<h3>How to identify the quartiles?</h3>

Divide the graph into 4 and analyze how much each quartile represents.

Learn more about graphs in: brainly.com/question/16608196

#SPJ1

5 0
2 years ago
(15 points) (Asap!!)<br>In what two ways can you increase the elastic potential energy of a spring?
adelina 88 [10]
Hello There

Answers: T<span>he elastic potential energy can be increased by: </span>

<span>1) Getting a spring with a higher spring constant</span>

<span>2) Increasing the length at which the spring is compressed. 

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris</span>
5 0
3 years ago
Read 2 more answers
A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be
vodomira [7]

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d  x C )

So electric field remains unchanged .

iv)

energy stored in first case = Q² / 2C

In the second case energy stored = Q² / 2x2C

so energy stored becomes half .

4 0
3 years ago
A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i
sweet [91]
<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V =  I x R                -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R               -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance.  Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

6 0
4 years ago
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