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Yuliya22 [10]
4 years ago
6

The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind

Physics
1 answer:
denis23 [38]4 years ago
6 0
Meteorites is the answer

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Describe, using the relevant physics, how moving a magnet near a [ 1 2 ] solenoid induces a voltage across it. How does the spee
Svetllana [295]

Answer:

Explanation:

Moving a magnet might cause a change in the magnetic field going through the solenoid. Whether or not it will change depends on the movement.

According to Faraday's law of induction a voltage is induced in a coil by a change in the magnetic flux. Magnetic flux is defined as the dot product of the magnetic field (a vector field) by the area enclosed by a loop of the coil.

\Phi B = -\int{B} \, dA

The voltage is induced by the variation of the magnetic flux:

\epsilon = -N * \frac{d \Phi B}{dt}

Where

ε: electromotive fore

N: number of turns in the coil

ΦB: magnetic flux

Moving the magnet faster would increase the rare of change of the magnetic flux, resulting in higher induced voltage.

Turning the magnet upside down would invert the direction of the magnetic field, reversing the voltage induced.

5 0
3 years ago
You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic
Alinara [238K]

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

5 0
3 years ago
A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this
mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0
3 years ago
1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms.
lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

4 0
3 years ago
Read 2 more answers
A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea
Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

4 0
2 years ago
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