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Dahasolnce [82]
2 years ago
14

We find that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared

to R, the radius of the disk. To see a numerical instance of this, calculate the magnitude of the electric field a distance y = 2.9 m from a disk of radius R = 2.9 cm that has a total charge of 5.0 µC using the exact formula as follows. (Enter your answer to at least one decimal place.)
E = 2πK σ(1- [y/(R² + y²)]) j^
Physics
1 answer:
Alex2 years ago
8 0

Answer:

7. 01 * 10^7 N/C

Explanation:

Parameters given:

Distance, y = 2.9 m

Radius, R = 2.9 cm = 0.029m

Charge, Q = 5.0 µC = 5 * 10^(-6) C

Given that:

E = 2πKσ(1- [y/(R² + y²)]) j^

Charge density, σ, is given as:

σ = Q/A = Q/πR²

=> E = 2πKQ/πR² (1- [y/(R² + y²)]) j^

E = 2KQ/R² (1 - [y/(R² + y²)]) j^

E = (2 * 9 * 10^9 * 5 * 10^(-6) / 0.029²) (1 - [2.9/(0.029² + 2.9²)]) j^

E = 107015.46 * 10^3 *(1 - 2.9/8.41) j^

E = 107015.46 * 10^3 * 0.655 j^

E = 7.01 * 10^7 j^ N/C

The magnitude of the electric field is 7.01 * 10^7 N/C. The j^ shows the direction.

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What travels around the earth but stays in one spot?
mina [271]
The sun?? It stays in one spot, but from our point of view, it travels around the earth...
7 0
3 years ago
A capoeira é muito mais que uma luta, por isso os campeonatos internos tem o objetivo de: 1 ponto Testar a força dos participant
timurjin [86]

Answer:

Avaliar o indivíduo como um todo.

Explanation:

O graduado na capoeira, é uma pessoa que recebe mais respeito pelo grupo. Sendo portanto o objetivo de cada exame interno, ou pelo menos deveria ser em sua vasta maioria avaliar o indivíduo como um todo.

8 0
2 years ago
A stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s. What is the frequency perceived by
inn [45]

For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived  is mathematically given as'

F=81.721Hz

<h3>What is the frequency perceived by a firefighter racing toward the station at 11km/h?</h3>

Generally, the equation for the doppler effect  is mathematically given as

F'=\frac{vs+v}{vs}*f

Therefore

F=81(343+3.05556)/343

F=81.721Hz

In conclusion, the frequency is

F=81.721Hz

Read more about frequency

brainly.com/question/24623209  

4 0
1 year ago
A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B f
Misha Larkins [42]

Answer:

A) False

B) False

C) True

D) False

Explanation:

A) False, because when leaving the field, the coil experiences a magnetic force to the right.

B) When the loop is entering the field, the magnetic flux through it will increase. Thus, induced magnetic field will try to decrease the magnetic flux i.e. the induced magnetic field will be opposite to the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is out of the plane of figure. Due to that reason, the current would be counterclockwise. So the statement is false.

C) When the loop is leaving the field, the magnetic flux through the loop will decrease. Thus, induced magnetic field will try to increase the magnetic flux i.e. the inducued magnetic field will be in the same direction as the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is also into the plane of figure. Due to that reason, the current would be clockwise. So the statement is true.

D) False because when entering the field magnetic force will be toward left side

8 0
3 years ago
One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t &lt; 5.0 s P(t)=60 W, 0≤
OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

P(t)=60 W, 0\leq t

P(t)=25 W, 5.0\leq t

Aplying the values to the equation we have:

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}

Power_{avg} = 42.5W

5 0
2 years ago
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