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3 years ago

14

Which configuration would produce an electric current? A) Rotate a coil of copper wire between two magnets. B) Connect a wire be

tween a copper and zinc strip sitting in a beaker of water. C) Connect a wire to the (+) positive end of a battery and the other end to a light bulb. D) Connect a wire to the (-) negative end of a battery and the other end to a light bulb.

2 answers:

Lelu [443]3 years ago

8 0

Answer:

A.) Rotate a coil of copper wire between two magnets.

Explanation:

Rotating a coil of copper wire between two magnets would produce an electric current. This is an example of electromagnetic induction.

FrozenT [24]3 years ago

4 0

the answer is (A) the movement of the magnet relative to the coil

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A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and

marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

$\lambda = \frac{v}{f}$

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

$\lambda = \frac{192}{320}$

$\lambda = 0.6m$

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

$N = \frac{L}{\lambda}$

$N = \frac{3.6}{0.6}$

$N = 6$

Therefore the number of wavelengths of the wave fit on the string is 6.

5 0

3 years ago

What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 10v/m?

Nitella [24]

Answer:

$3.33\cdot 10^{-8} T$

Explanation:

For an electromagnetic wave, the relationship between magnetic field amplitude and electric field amplitude is given by

$E=cB$

where

E is the amplitude of the electric field

c is the speed of light

B is the amplitude of the magnetic field

For the electromagnetic wave in this problem, we have

E = 10 V/m is the amplitude of the electric field

So if we solve the formula for B, we find the amplitude of the magnetic field:

$B=\frac{E}{c}=\frac{10 V/m}{3\cdot 10^8 m/s}=3.33\cdot 10^{-8} T$

4 0

3 years ago

A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i

Lerok [7]

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

5 0

2 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

6. What is the change in temperature of a metal rod that is 55.0 cm long, decreases length by 0.20 cm, and that has a coefficien

MrMuchimi

Explanation:

We have,

Length of a metal rod is 55 cm or 0.55 m

Change in length is 0.2 cm or 0.002 m

It is required to find the change in temperature of a metal rod. The coefficient of linear expansion is given by :

$\alpha =\dfrac{\Delta L}{L_0\Delta T}$

$\Delta T$ is the change in temperature

$\Delta T =\dfrac{\Delta L}{L_0\alpha }\\\\\Delta T =\dfrac{0.002}{0.55\times 12\times 10^{-6}}\\\\\Delta T= 303.03^{\circ} C$

So, the change in temperature is 303.03 degrees Celsius.

4 0

3 years ago