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Sophie [7]
3 years ago
12

Which of the following types of electromagnetic radiation has the most energy?

Physics
2 answers:
Thepotemich [5.8K]3 years ago
6 0

The ultraviolet is the answer!

marysya [2.9K]3 years ago
4 0
You didn't post any choices. A microwave Photon carries more energy than a radio wave photon. An infrared Photon carries more energy than a microwave photon. A visible light Photon carries more energy than an infrared photon. An UltraViolet Photon carries more energy than a visible light photon. An x-ray Photon carries more energy than an UltraViolet photon. A gamma ray Photon carries more energy than an x-ray photon.
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A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
3 years ago
Read 2 more answers
Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions? ​
erica [24]

Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

8 0
3 years ago
In the following figure, electric field at y axis will be maximum at y=?
Strike441 [17]

Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌

5 0
3 years ago
Read 2 more answers
What equals the number of protons in an atom
Reptile [31]

<span>Each atom contains an equal number of protons and electrons; these particles will be equal in value to an element's atomic number</span>

6 0
3 years ago
Read 2 more answers
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

Therefore, the number of turns of wire needed is 573.8 turns

4 0
3 years ago
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