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faltersainse [42]
4 years ago
14

A circular loop of wire is positioned so the normal to its area points vertically upward. A bar magnet is dropped from above the

loop so that its north pole falls through the loop first. When looking at the loop from above, in which direction is the induced current in the loop just as the north pole of the magnet enters the loop?
a. Clockwise
b. There is no induced current Incorrect.
Physics
1 answer:
Kruka [31]4 years ago
8 0
B cause there is no induced current incorrect
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A 0.5 kg stone is raised from 1m to 2m height from the ground. what is the change in potential energy of the stone?
Usimov [2.4K]

Given: The mass of stone (m) = 0.5 kg

Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m

Acceleration due to gravity (g) = 9.8 m/s²

To find: The change in potential energy of the stone

Formula: The potential energy (P) = mgh

where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

Δ P = P₂ - P₁ = mg (h₂ - h₁)

or,                = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)

or,                = 4.9 J

Hence, the required change in the potential energy of the stone will be 4.9 J

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3 years ago
What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?
Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

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6 0
4 years ago
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Which best describes the relationship between the direction of energy and wave motion in a transverse wave?
sammy [17]
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Deanna stirred a teaspoon of sugar into a glass of warm water. The sugar completely dissolved in the water. Select 3 statements
larisa [96]

Answer:

B, C, F

Explanation:

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C: Sugar gets spread out among the water.

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8 0
3 years ago
Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state
topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

<u>Given  </u>

<u><em>Process 1 ---> 2 </em></u>

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

Volume of point (1) V1   = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant  

<u>Process 2 ---> 3 </u>

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

<u>Process 3 ---> 1 </u>

The relation of the process PV = constant  

<u>Required  </u>

Sketch the processes on the PV coordinates

The work for each process in kJ  

<u>Solution  </u>

The work is defined by  

W=\int\limits^a_b {x} \, dx

<em>a=V2</em>

<em>b=V1</em>

<em>x=P</em>

<em>dx=dV</em>

<u>Process 1 ---> 2  </u>

P3 = P4 = 5 bar  

W=\int\limits^a_b {x} \, dx

<em>a=V3</em>

<em>b=V2</em>

<em>x=4</em>

<em>dx=dV</em>

putting the value of a, b, x, dx in above integral

W=400 kJ

<u>Process 2 ---> 3 </u>

V = constant Then there is no change in the volume,hence W = 0 kJ  

<u>Process 3 ---> 1  </u>

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1  

 W=\int\limits^a_b {x} \, dx

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

  = -160.944 KJ

5 0
3 years ago
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