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nevsk [136]
3 years ago
7

Assignment 1B-X.P.6 lesson 1

Physics
1 answer:
pshichka [43]3 years ago
3 0

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Which of the following choices correctly identifies the process created by moving a magnet through a conducting loop?
jolli1 [7]

Answer: D.) electromagnetic induction

Explanation: Electroctromagnetic induction may be explained as a process whereby electric current is induced or produced by difference in potential resulting from the movement of conductor across a magnetic field.

In simple terms, an electromotive force is induced when a magnet is moved through a conducting loop.

The electromotive force produced by moving a magnet through a conducting loop can be represented by the relation:

E = - N (dΦ / dt)

Where E = electromotive force in voltage

N = number of loop in conductor

dΦ = change in magnetic Flux

dt = change in time

3 0
3 years ago
Theories have both an explanatory and a predictive function. <br> a. True <br> b. False
Simora [160]
Theories have both an explanatory an a predictive function. True
3 0
3 years ago
Read 2 more answers
The Venus flytrap is known for which of these behaviors?
tensa zangetsu [6.8K]
Snapping a leaf shut around an insect, I think.
8 0
2 years ago
The circuit below shows some of the circuity in a small toy robot. When the circuit is on, the robot moves its arms (the motor)
djyliett [7]

Answer:

dear can you provide the circuit of the robot

6 0
3 years ago
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the
Julli [10]

Answer:

<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0
3 years ago
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