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3 years ago

13

Suppose you were asked to demonstrate electromagnetic induction. Which of the following situations will result in an electric cu

rrent?
A. A magnet is moving toward a wire loop.
B. A wire loop is moving away from a magnet.
C. A wire loop is rotated near a magnet.
D. All of the above

2 answers:

steposvetlana [31]3 years ago

5 0

D. All of the above. When a wire loop is moved or rotated in a magnetic field, there is a change in magnetic flux which produces emf in wire loop and hence electric current is produced.

ad-work [718]3 years ago

3 0

The answer is D. It is obvious since it is a one-choice question and both A and B state the exact same thing. Therefore the answer has to be D and so C has to be correct as well. Exam techniques! :D

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

3 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

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3 years ago

What is the mass of basswood wing with these dimensions: 26.9 cm x 5.5 cm x 0.15 cm?

Softa [21]

The answer is 22.1925 I hope you get right let me know I got my answer because I multiply it all together.

6 0

3 years ago

Read 2 more answers

Which of these events indicate that a chemical reaction might have occurred?

blagie [28]

Reactions occur when two or more molecules interact and the molecules change. Bonds between atoms are broken and created to form new molecules. That's it.

7 0

3 years ago

The lower the__ the frequency, the __the pitch <br><br> -lower <br> -higher

rjkz [21]

Answer:

umm the lower the frequency the higher the pitch

Explanation:

6 0

3 years ago