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erica [24]
3 years ago
15

Find an equation for the plane that passes through the point [1, 1, 8] and that has [-1, -6, 3] as normal vector.

Physics
1 answer:
OlgaM077 [116]3 years ago
3 0
The normal to the plane is (Fx, Fy, Fz) = (1, 1, 8)
We have (x0, y0, z0) = (-1, - 6, 3)

Equation of the plane:
Fx(x - x0) + Fy(y - y0) + Fz(z - z0) = 0
=> (1)(x + 1) + (1)(y + 6) + (8)(z - 3) = 0
=> x + 1 + y + 6 + 8z - 24 = 0
=> x + y + 8z = 17
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A solid ball of radius rb has a uniform charge density rho.
Oksana_A [137]

Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a.  What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?  Express your answer in terms of ρ, rb, r, and ϵ0.

b.   What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?  Express your answer in terms of ρ, r, rb, and ϵ0.

c.   Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ ,  ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V=  ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E=  ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b)  The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E=  ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

(c)  E(0)= 0

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

4 0
3 years ago
Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022
monitta

Answer:

F=1.95\times 10^{20}\ N

Explanation:

Mass of Earth, m_e=5.98 \times 10^{24}\ kg

Mass of Moon, m_m=7.26\times 10^{22}\ kg

The distance between Earth and the Moon is, d=384,400\ km

We need to find the force of gravitational attraction between the Earth and the moon. The force of gravity is given by :

F=G\dfrac{m_em_m}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{5.98 \times 10^{24}\times 7.26\times 10^{22}}{(384400 \times 10^3)^2}\\\\F=1.95\times 10^{20}\ N

So, the required force is 1.95\times 10^{20}\ N.

4 0
3 years ago
(15 points) (Asap!!)<br>In what two ways can you increase the elastic potential energy of a spring?
adelina 88 [10]
Hello There

Answers: T<span>he elastic potential energy can be increased by: </span>

<span>1) Getting a spring with a higher spring constant</span>

<span>2) Increasing the length at which the spring is compressed. 

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris</span>
5 0
3 years ago
Read 2 more answers
Which changes in energy form are illustrated in the diagram.
Grace [21]
It would be B since it starts with the solar energy which is converted to electricity with the solar panels, which then creates mechanical energy for the fans blades to move and sound for the radio.

Hope that helps :)
6 0
3 years ago
A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4
Vedmedyk [2.9K]
Hello!

Vx = V0x + Ax*t
Vx = 18.1 + 2.4t

Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s

Then, the final velocity of the car is 36.1 m/s.
3 0
3 years ago
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