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vitfil [10]
3 years ago
9

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

point?
Express your answer to two significant figures and include the appropriate units.

Part B) If a 20 nC charge were placed at this point, what would its electric potential energy be?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Anastasy [175]3 years ago
5 0

Answer:

Part A:

\rm 2.8\times 10^3\ Volts.

Part B:

\rm 5.6\times 10^{-5}\ J.

Explanation:

<u> Part A:</u>

  • Potential energy of charge at the given point, \rm U=42\ \mu J=42\times 10^{-6}\ J.
  • Charge, \rm q=15\ nC = 15\times 10^{-9}\ C.

The potential energy at a point due to a charge is defined as

\rm U=qV.

<em>where</em>,

V = electric potential at that point.

Therefore,

\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.

<u>Part B:</u>

Now, if the charge at that point is replaced with \rm q_1 = 20\ nC = 20\times 10^{-9}\ C., then the electric potential energy at that point is given by

\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.

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A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0
3 years ago
Help me please and thank you
katovenus [111]

Answer:

go to : www.planetresourses.com/test2.00/answers, ant type in that test name

Explanation:

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3 0
3 years ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>

<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
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4 0
3 years ago
What is the power of a machine that pushes with a force of 3 n for a distance of 9 m in 8 s?
dalvyx [7]
The work done by the machine is equal to the product between the force applied and the distance over which the force is applieds, so in this case:
W=Fd=(3 N)(9 m)=27 J

And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
P= \frac{W}{t}= \frac{27 J}{8 s}=3.38 W

4 0
3 years ago
Question: Point B is 25 km due east of point A. Starting from point A, a camel walks 20 km in
Korolek [52]

<em>Resultant angle; θ = 25.59°  </em>

This question is dealing with bearings and distance.

We are told that from point A, the camel walks 20 km at 15° in the south of east direction.

Thus, d_s,e = 20 km

Resolving along the horizontal east direction gives; d_e = 20 cos 15

d_e = 19.32 km

Also, resolving along the vertical south direction gives; d_s = 20 sin 15

d_s = 5.18 km

Net vertical distance; d_vert = 8km - 5.18km = 2.72 km

Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km

Now, the resultant angle is given by;

tan θ = d_vert/d_hor

tan θ = 2.72/5.68

tan θ = 0.4789

θ = tan^(-1) 0.4789

θ = 25.59°

Read more at; brainly.com/question/22518031

8 0
3 years ago
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