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givi [52]
3 years ago
10

What do you think we call this graphical representation based on your prior experience with electric fields and electric field l

ines? Is this quantity (which you are observing though the compass's behavior) a vector or scalar? How do you know?
Physics
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

Explanation:

The strengthcompassion field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. The direction of the electric field is tangent to the field line at any point in space. Field lines can never cross. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges.

Rules for drawing electric field lines

1. Electric field lines are always drawn from High potential to

low potential.

2. Two electric field lines can never intersect each other.

3. The net electric field inside a Conductor is Zero.

4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.

5. The density of electric field lines tells the strength of the electric field at that region.

6. Electric field lines terminate Perpendicularly to the surface of a conductor.

A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.

So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.

Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity

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The fuzzy cloud around the nucleus is called the
Kamila [148]
It's number three on your worksheet. ;)
6 0
3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
4 years ago
A stationary boat in the ocean is experiencing waves from a storm. the waves move at 56 km/h and have a wavelength of 160 m, bot
Lera25 [3.4K]
The wavelength \lambda of the wave is 160 m, and this is the distance between two consecutive crests. The boat is located at a crest of the wave, this means that the first trough is located 80 meters from the boat (because the distance between a crest and a trough is half the wavelength).

The speed of the wave is 
v=56 km/h = 15.6 m/s
so the time the boat takes to reach the first trough is
t= \frac{S}{v} = \frac{80 m}{15.6 m/s}=5.1 s
5 0
4 years ago
The period of a sound wave coming from an instrument is 0. 002 seconds. What is the frequency of the sound? Hz.
Pie

The period is the time taken by the wave to complete an oscillation. The frequency of the given sound is 500 Hz.

<h2>Period:</h2>

It is the time taken by the wave to complete an oscillation. The frequency is inversely proportional to the time:

f = \dfrac 1T

Where,

f- frequency

T - period = 0.002 s

Put the value in the equation,

f = \dfrac 1{0.002}\\\\f = 500\rm \  Hz

Therefore, the frequency of the given sound is 500 Hz.

Learn more about Period:

brainly.com/question/842349

5 0
2 years ago
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