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Andrej [43]
3 years ago
9

Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-

µF capacitor when it is fully charged?
Physics
1 answer:
torisob [31]3 years ago
5 0

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

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The theory of plate tectonics describes how the Earth’s lithosphere is broken into plates and these plates move over the molten
Mumz [18]

Answer:

C

Explanation:

A theory can be changed when new evidence is found. A law doesn't change because it is universally a fact. It doesn't need new evidence to support it.

5 0
3 years ago
How do you calculate the net force, i need a full explanation PLEASE
Lina20 [59]

Answer:

Once you have drawn the free-body diagram, you can use vector addition to find the net force acting on the object. We will consider three cases as we explore this idea:

Case 1: All forces lie on the same line.

If all of the forces lie on the same line (pointing left and right only, or up and down only, for example), determining the net force is as straightforward as adding the magnitudes of the forces in the positive direction, and subtracting off the magnitudes of the forces in the negative direction. (If two forces are equal and opposite, as is the case with the book resting on the table, the net force = 0)

Example: Consider a 1-kg ball falling due to gravity, experiencing an air resistance force of 5 N. There is a downward force on it due to gravity of 1 kg × 9.8 m/s2 = 9.8 N, and an upward force of 5 N. If we use the convention that up is positive, then the net force is 5 N - 9.8 N = -4.8 N, indicating a net force of 4.8 N in the downward direction.

Case 2: All forces lie on perpendicular axes and add to 0 along one axis.

In this case, due to forces adding to 0 in one direction, we only need to focus on the perpendicular direction when determining the net force. (Though knowledge that the forces in the first direction add to 0 can sometimes give us information about the forces in the perpendicular direction, such as when determining frictional forces in terms of the normal force magnitude.)

Example: A 0.25-kg toy car is pushed across the floor with a 3-N force acting to the right. A 2-N force of friction acts to oppose this motion. Note that gravity also acts downward on this car with a force of 0.25 kg × 9.8 m/s2= 2.45 N, and a normal force acts upward, also with 2.45 N. (How do we know this? Because there is no change in motion in the vertical direction as the car is pushed across the floor, hence the net force in the vertical direction must be 0.) This makes everything simplify to the one-dimensional case because the only forces that don’t cancel out are all along one direction. The net force on the car is then 3 N - 2 N = 1 N to the right.

Case 3: All forces are not confined to a line and do not lie on perpendicular axes.

If we know what direction the acceleration will be in, we will choose a coordinate system where that direction lies on the positive x-axis or the positive y-axis. From there, we break each force vector into x- and y-components. Since motion in one direction is constant, the sum of the forces in that direction must be 0. The forces in the other direction are then the only contributors to the net force and this case has reduced to Case 2.

If we do not know what direction the acceleration will be in, we can choose any Cartesian coordinate system, though it is usually most convenient to choose one in which one or more of the forces lie on an axis. Break each force vector into x- and y-components. Determine the net force in the x direction and the net force in the y direction separately. The result gives the x- and y-coordinates of the net force.

Example: A 0.25-kg car rolls without friction down a 30-degree incline due to gravity.

We will use a coordinate system aligned with the ramp as shown. The free-body diagram consists of gravity acting straight down and the normal force acting perpendicular to the surface.

We must break the gravitational force in to x- and y-components, which gives:

F_{gx} = F_g\sin(\theta)\\ F_{gy} = F_g\cos(\theta)F

gx

​

=F

g

​

sin(θ)

F

gy

​

=F

g

​

cos(θ)

Since motion in the y direction is constant, we know that the net force in the y direction must be 0:

F_N - F_{gy} = 0F

N

​

−F

gy

​

=0

(Note: This equation allows us to determine the magnitude of the normal force.)

In the x direction, the only force is Fgx, hence:

F_{net} = F_{gx} = F_g\sin(\theta) = mg\sin(\theta) = 0.25\times9.8\times\sin(30) = 1.23 \text{ N}F

net

​

=F

gx

​

=F

g

​

sin(θ)=mgsin(θ)=0.25×9.8×sin(30)=1.23 N

7 0
3 years ago
A block of mass 0.254 kg is placed on top of a light, vertical spring of force constant 5 100 N/m and pushed downward so that th
ryzh [129]

Answer:

8.86 m

Explanation:

According to the law of conservation of energy, the elastic potential energy initially stored in the spring will be converted into gravitational potential energy of the block when it is at its maximum height:

\frac{1}{2}kx^2 = mgh

where

k = 5100 N/m is the spring constant

x = 0.093 m is the spring compression

m = 0.254 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

h is the maximum height of the block

Solving the equation for h, we find

h=\frac{kx^2}{2mg}=\frac{(5100 N/m)(0.093 m)^2}{2(0.254 kg)(9.8 m/s^2)}=8.86 m

3 0
3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 1.25 s at the same height as the point of release.
Alenkasestr [34]

Answer:

A. 9.8 m/s²

B. Zero

C. 6.125 m/s

D. 1.91 m

Explanation:

From the question given above, the following data were obtained:

Time (T) spent in the air = 1.25 s

A. Determination of the acceleration of the ball.

From the description given in question above, the motion of the tennis ball is motion under gravity. Hence, the ball will experience an acceleration due to gravity of 9.8 m/s²

B. Determination of the velocity at maximum height.

Maximum height is the greatest point reached by the tennis ball above the ground. At maximum height, the velocity of the tennis ball is zero since it has no further force to propel it upward.

C. Determination of the initial velocity of the ball.

We'll begin by calculating the time taken to reach the maximum height. This can be obtained as follow:

Time (T) spent in the air = 1.25 s

Time (t) to reach the maximum height =?

T = 2t

1.25 = 2t

Divide both side by 2

t = 1.25 / 2

t = 0.625 s

Finally, we shall determine the initial velocity of the ball. This can be obtained as follow:

Time (t) to reach the maximum height = 0.625 s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 0.625)

0 = u – 6.125

Collect like terms

0 + 6.125 = u

u = 6.125 m/s

Thus, the initial velocity of the ball is 6.125 m/s

D. Determination of the maximum height.

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) = 6.125 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 6.125² – (2 × 9.8 × h)

0 = 37.52 – 19.6h

Collect like terms

0 – 37.52 = – 19.6h

– 37.52 = – 19.6h

Divide both side by – 19.6

h = – 37.52 / – 19.6

h = 1.91 m

Thus, the maximum height reached by the ball is 1.91 m

3 0
3 years ago
Write a function to accept a vector of masses (m) from the user and gives the corresponding energy to them. Energy vector is the
jarptica [38.1K]

Answer:

Written in Python

def energyvector(mass):

    c = 2.9979 * 10**8

    energy = mass * c ** 2

    print(round(energy,2))

Explanation:

This line defines the function

def energyvector(mass):

This line initializes the speed of light

    c = 2.9979 * 10**8

This line calculates the corresponding energy

    energy = mass * c ** 2

This line prints the calculated energy

    print(round(energy,2))

7 0
3 years ago
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