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kondaur [170]
4 years ago
7

You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic

h of the following increases the energy density? (Select all that apply.)
a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed
b. increasing the number of turns per unit length on the solenold
c. increasing the cross-sectional area of the solenoid
d. none of these
e. increasing the current in the solenoid
Physics
1 answer:
Alinara [238K]4 years ago
5 0

Answer:

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

Explanation:

As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As

B = μ₀nI

n = no of turns

I = current through the wire

So the right options are

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

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In figure 1, charge q2 experiences no net electric force. What is q1?
lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

F = k\frac{q_1*q_2}{r^2}

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}

And we know that it must be equal to zero, so we can write it as:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\

The parenthesis must be equal to zero, so we can write:

\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0

And now we can solve this for q₁ to get:

q_1  = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0
3 years ago
What is the wavelength of a wave that is traveling at 343 m/s with a frequency of 1000 Hz
ZanzabumX [31]

Answer:

343 \div 1000 = 0.343

wavelength is equivalent to the velocity divided by the frequency

3 0
3 years ago
When magnesium ribbon is ignited, it gives off a white fume of magnesium oxide. What is this type of reaction?
Scilla [17]
It is a synthesis reaction where two reactants gives one product.

Answer:option D

2Mg + O2 ------> 2 MgO
4 0
3 years ago
Read 2 more answers
A power plant taps steam superheated by geothermal energy to 475 K (the temperature of the hot reservoir) and uses the steam to
jasenka [17]

Answer:

Thermal Efficiency, η = \frac{W₀}{Q₁}   . . . . . . . . . . . . . . . . Eqn 1

where W₀ = Work Output = Q₁ - Q₀ =82500KW    . . . . . . .    . . . . . . . . Eqn 2

Q₁ = Heat Supplied/Input = mC(ΔT₁)

Q₁ = Heat Rejected/Output = mC(ΔT₀)     . . . . . . . . . . . . . . . . . . . . . . . . Eqn 3

Note:  From Carnot's theorem, for any engine working between these two temperatures (T₀/T₁), The maximum attainable efficiency is the Carnot efficiency given as follows;

Therefore, η = 1 - \frac{Q₀}{Q₁} = 1 - \frac{T₀}{T₁}

Remember, T₁ = 475K and T₀ = 308K

η = 1 - (308/475) = 1 - 0.648 = 0.352

Hence, the maximum efficiency at which this plant can operate = 35%

2. To determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

Remember from Eqn 1, Q₁ = W/η,

Therefore, Q₁=  82500/0.35

  Q₁=235,714KW,

So, from Eqn 2, Q₀ = 235714 - 82500

                                Q₀ = 153214KW (KJ/s)  (Released Heat)

In t =24 hours, we can then use this to determine the minimum amount of heat rejected qₓ (KiloJoule),  = Q₀t  (Remember, you have to convert the time, t, unit to seconds)

                                           = 153214 x t (KiloJoule)

qₓ = 153214 x 24 x 3600 (KiloJoule)

qₓ = 13238 MegaJoule

<h3>Therefore, the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours is 13238 MJoule</h3>
4 0
3 years ago
Select an incorrect statement relating to a Carnot cycle operating on a gas. A) Each process is a reversible process. B) work oc
nekit [7.7K]

Answer:

B)

D)

Explanation:

Carnot cycle ;

  This is the ideal cycle for all type of heat engine.All the process in the Carnot cycle is reversible processes ( those process does not leave any effect on the system as well as surrounding when the process is  reversed  is known as reversible process).It have four process in which two process is constant temperature and other two is isentropic or we can say that reversible adiabatic.

In the two process(constant temperature) only heat interaction take places and other two process(adiabatic) only work interaction take places.

So the option B and D is incorrect for Carnot cycle.

4 0
3 years ago
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