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FromTheMoon [43]
3 years ago
5

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr

ical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

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jonny [76]

Answer:

2 m/s

Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.

Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.

Velocity = (8 - 6) m/s

Velocity = 2 m/s

Therefore, the jogger will be moving at the speed of 2 m/s

5 0
3 years ago
A. Draw the electric field lines around a negative charge.
Alborosie
<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric  fields, so this helps us to see the problem more real. So an electric field line is an imaginary  line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional  to the product of the charges and inversely proportional to the square  of the distance between them. </em>This can be expressed as follows:

F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m

Then:

F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge  and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

q_{1} \ and \ q_{2}

If the amount of charge on one of the objects is tripled, let's say this is the charge q_{2}, then the new charge is:

q_{N}=3q_{2}

In the formula of Coulomb:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2}

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod  repels positive charges on the sphere, creating  zones of negative and  positive charge as indicated in the second Figure.

7 0
3 years ago
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labwork [276]

Answer/Explanation:

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acceleration = v - v0 / t

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8 0
3 years ago
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Which of the following is NOT a vector quantity?
lora16 [44]

Answer:

D

Explanation:

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average velocity is vector

linear momentum = mass× velocity

therefore it's also vector

force = mass× acceleration

acceleration is vector therefore force also vector

potential energy= mgh

energy in any form is scaler quantity therefore potential energy too is a scaler quantity.

5 0
3 years ago
The center of mass of a 0.3 kg (non-uniform) meter stick is located at its 46-cm mark. What is the magnitude of the torque (in N
Sati [7]

Answer:

15cm

Explanation:

7 0
3 years ago
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