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exis [7]
2 years ago
11

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

f consecutive lines is constant. Clear the equipotential lines using the Erase button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don't worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines?
1. The equipotential lines are closer together in regions where the electric field is weaker.
2. The equipotential lines are closer together in regions where the electric field is stronger.
3. The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Physics
1 answer:
Elza [17]2 years ago
5 0

Answer:

B or 2

Explanation:

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1. The velocity decreases, and the kinetic energy decreases.

2. An increase in temperature difference between the inside and outside of the building.

3. The total kinetic energy remains the same.

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2. What is the percent composition of sulfur in H2SO4?
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What is the most appropriate tool to use when collecting data from the investigation? List, table, graph or chart?
forsale [732]
A table would be the most appropriate because that way you can compare the data.
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3 years ago
The speed of light in a vacuum is approximately 2.99 × 108 m/s, and the speed of light through a piece of glass is approximately
exis [7]

Option (b) is correct.The index of refraction for the glass is 1.52

Explanation:

velocity of light in vacuum= C= 2.99 x 10⁸m/s

Velocity of light in glass = V= 1.97 x 10⁸m/s

The refractive index is given by n=\frac{c}{v}

n= 2.99 x 10⁸/1.97 x 10⁸m/s

n= 1.52

Thus the refractive index of glass is 1.52

6 0
3 years ago
Read 2 more answers
The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

4 0
3 years ago
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