Typical noise associated with failed cv joint

Pls help me it’s due today

hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

6 0

2 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1

WITCHER [35]

Answer:

Detailed solution is given below:

7 0

3 years ago

Sometimes, steel studs may not be used on outside walls because they are?

Helen [10]

Answer:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Explanation:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

6 0

3 years ago

16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

See full question below

You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

Learn more about two-way road from

brainly.com/question/13123201

#SPJ2

5 0

1 year ago

Typical noise associated with failed cv joint​

Typical noise associated with failed cv joint