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2 years ago

11

The energy of a certain charged capacitor is 7 J. What is the new energy stored in that capacitor if its charge is decreased to

1 / 4 of its original value (imagine allowing some of the charge to move through a resistor)? Remember that the capacitance, C , that relates Q and V is unchanged.
Enew=?

1 answer:

poizon [28]2 years ago

6 0

Answer:

The new energy is 1/16 of the original energy

Explanation:

A capacitor is an electric device which is able to store charge when it is connected to a power supply.

The energy stored in a capacitor is given by the equation

$E=\frac{Q}{2C}$

where:

Q is the charge stored on the capacitor

C is the capacitance of the capacitor

For the capacitor in this problem, initially we have

E = 7 J (energy) when the charge stored is Q and the capacitance is C

Later, the charge is decreased to 1/4 of its original value, so the new charge is

$Q'=\frac{1}{4}Q$

Since the capacitance remains the same, the new energy is

$E'=\frac{(\frac{1}{4}Q)^2}{2C}=\frac{1}{16}\frac{Q^2}{2C}=\frac{1}{16}E$

Therefore, the new energy is 1/16 of the original energy.

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(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c

umka2103 [35]

Answer:

$E=\frac{k\,Q}{d^2}$

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: $E=\frac{k\,Q}{d^2}$ , where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

$F_C=k\frac{Q_1\,Q_2}{d^2}$

but modified with only one charge showing in the numerator of the expression.

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2 years ago

A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

$v = 70 m/s$

now we have

$v_x = 70 cos25$

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$v_y = 70 sin25$

$v_y = 29.6 m/s$

now in Y direction we can use kinematics

$v_y = v_i + at$

$v_y = 29.6 - (9.81 \times 5)$

$v_y = -19.5 m/s$

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{63.44^2 + 19.5^2}$

$v = 66.4 m/s$

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2 years ago

Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th

kondor19780726 [428]

When plane is going towards Halifax the speed of wind is in the direction of fly

so overall the net speed of the plane will increase

while when he is on the way back the air is opposite to flight so net speed will decrease

now the total time of the journey is 13 hours

out of this 2 hours he spent in mathematics talk

so total time of the fly is 13 - 2 = 11 hours

now we have formula to find the time to travel to Halinex

$t_1 = \frac{d}{v + 50}$

time taken to reach back

$t_2 = \frac{d}{v - 50}$

now we have total time

$T = t_1 + t_2$

$11 = \frac{d}{v - 50} + \frac{d}{v + 50}$

here d= 3000 miles

$11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}$

$3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}$

$v^2 - 2500 = 545.45v$

solving above quadratic equation we will have

$v = 550 mph$

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3 0

2 years ago

Read 2 more answers

This means that the speed at which the bullet travels across Earth's surface (its magnitude of horizontal velocity) does not aff

Dmitry_Shevchenko [17]

Answer: the speed at which it falls toward the Earth.

Explanation:

A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.

Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).

This kind of motion is the result of two independent motions: vertical motion and horizontal motion.

The observed net velocity is the vectorial sum of the vertical and horizontal velocities.

The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).

The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.

Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

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2 years ago

You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force

Tpy6a [65]

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

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a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

<u>Mass of Little Sister = 44.17 kg</u>

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2 years ago