Electric potential difference is the change in kinetic energy per unit charge in an electric field. Is this true or false?

The evaluation of the capacitor (in series) network is as follows;

Part A

Q = 3.2 μC

Part B

Q₁ = 3.2 μC

Part C

Q₂ = 3.2 μC

Part D

U = 76.8 μJ

Part E

U₁ = 34 2/15 μJ

Part F

U₂ = 53 1/3 μJ

Part G

V₁ = 21 1/3 V

Part H

V₂ = 26 2/3 V

A capacitor consists of pairs of conductors separated by insulators. Capacitors are used to store electric charge.

The specified parameters are;

The voltage across ab = 48 V

The capacitance of the first capacitor, C₁ = 150 nF

Capacitance of the second capacitor, C₂ = 120 nF

Part A

The total charge in a capacitor network can be found as follows;

\(C_{eq} = \left(\dfrac{1}{150} + \dfrac{1}{120} \right)^{-1} nF = \left(\dfrac{3}{200} \right)^{-1}nF\)

\(C_{eq} =\left(\dfrac{3}{200} \right)^{-1}nF=66\frac{2}{3} \, nF\)

\(Q_{eq} = C_{eq}\times V_{ab}\)

Therefore;

\(Q_{eq}\) = 66 2/3 nF × 48 V = 3,200 × 10⁻⁹ C = 3.2 μC

Part B

The charge in the 150 nF capacitor is obtained from the formula for the charge in a capacitor; Q = C × V as follows;

Q = C₁V₁ = C₂V₂

The charge in the capacitors, C₁ and C₂ are the same as the total charge of 3.2 μC

The charge, Q₁ on the 150 nF capacitor, C₁ is therefore, 3.2 nC

Part C

The capacitors, C₁ and C₂ are in series, therefore, the charge in each capacitor is equivalent to the charge in the circuit, which is 3.2 μC.

Therefore, the charge, Q₂, in the 120 nF capacitor, C₂ is 3.2 μC

Q₂ = 3.2 μF

Part D

The total energy stored in the network can be obtained using the formula;

U = (1/2)·C·V²

Where;

U = The energy in the capacitor

C = The equivalent capacitance of the network = 66 2/3 nF

V = The voltage

Therefore;

\(U = \dfrac{1}{2} \times C_{eq}\times V^2\)

\(U = \dfrac{1}{2} \times 66\frac{2}{3} \times 10^{-9}\times 48^2 = 76.8\)

Part E

The energy stored in the 150 nF capacitor is found as follows;

\(Q_{eq}\) = Q₁ = C₁ × V₁

V₁ = \(Q_{eq}\) ÷ C₁

Therefore;

V₁ = 3.2 μC ÷ 150 nF = \(21\frac{1}{3}\) V

U₁ = 0.5×C₁×V₁²

Part F

The energy stored in the 120 nF capacitor, U₂, can be found as follows;

V₂ = 3.2 μC ÷ 120 nF = \(26\frac{2}{3}\) V

U₂ = 0.5 × 150 nF × \(\left(26\frac{2}{3} \, V\right)^2\) = \(53\frac{1}{3}\, \mathrm{ \mu J}\)

Part G;

The potential difference across the 150 nF, obtained in Part E, is 21 1/3 V

Part H

The potential difference across the 120 nF, obtained in part F, is 26 2/3 V

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Answer:1.5 faster

Explanation:

The diameter of the base of a tapered drinking cup is 6 cm. The diameter of its mouth is 9 cm.

Answer:

the answer is weight

Explanation:

Answer:

Weight

Explanation: Just took the quiz

Therefore, the magnetic force on an electron is 1.175 x 10⁻¹⁴N in the positive y-direction.

The force experienced by a moving charge in a magnetic field is given by the Lorentz force law. Since the charge on a proton is positive and that on an electron is negative, the direction of the magnetic force experienced by each is different.In this question, we need to find the magnitude and direction of the magnetic force on a proton and an electron as they travel from the Sun to the Earth.

Let's first calculate the magnetic force on a proton:

F = qvBsinθ

where q = charge of the particle

v = velocity of the particle

B = magnetic field strength

θ = angle between the velocity of the particle and the magnetic field = 90° (since the proton is moving perpendicular to the magnetic field)

Therefore,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= 1.175 x 10⁻¹⁴ N

Direction of magnetic force on a proton:The direction of the magnetic force on a proton can be found using the right-hand rule. According to this rule, if we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the negative y-direction.

Therefore, the magnetic force on a proton is 1.175 x 10^-14 N in the negative y-direction.

- Now, let's calculate the magnetic force on an electron:

Again using the Lorentz force law,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= -1.175 x 10⁻¹⁴ N (the negative sign indicates that the direction of the magnetic force is opposite to that of the proton)

Direction of magnetic force on an electron:Again using the right-hand rule, we can find the direction of the magnetic force on an electron. If we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the positive y-direction.

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Answer: B

Explanation: B is the correct statement describing the positions of the Moon, the Sun, and the Earth during a new moon. The Moon is between Earth and the Sun.

During a new moon, the Moon is positioned between the Sun and the Earth, with the illuminated side of the Moon facing away from the Earth. This means that the side of the Moon that faces the Earth is not receiving any sunlight, making it invisible to us from Earth. The new moon is the first phase of the lunar cycle and occurs roughly every 29.5 days.

a. The image distance is -212.275 cm

b.  the lateral magnification is equal to 26.534375

The image distance can be found using the lens equation for a spherical lens:

1/f = 1/d_o + 1/d_i

we have that  focal length of a spherical lens is:

f = R / (n - 1)

f = R / (n - 1) = 200 cm / (1.52 - 1) = 200 cm / 0.52 = 385.385 cm

The object distance, d_o, is 8.00 cm,

calculating  the image distance, we have :

1/f = 1/d_o + 1/d_i

1/385.385 cm = 1/8.00 cm + 1/d_i

d_i = -8.00 cm * 385.385 cm / (385.385 cm - 8.00 cm) = -212.275 cm

b. The lateral magnification can be described as the ratio of the height of the image to the height of the object:

m = h_i / h_o

m = h_i / h_o = d_i / d_o = -212.275 cm / 8.00 cm = -26.534375

In conclusion, the negative sign indicates that the image is inverted with respect to the object.

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Answer:

1122.8

Explanation:

12.73 kg x 9.8 m/s^2 x 9m

=1122.786

Rounded=1122.8

The velocity gradient across the 3.7 mm thick oil layer on the flat plate sliding on the horizontal table is 0.594 m/s/m.

Given:

Surface area of the plate (A) = 1.8 m^2

Thickness of the oil layer (h) = 3.7 mm = 0.0037 m

Force applied on the plate (F) = 2.2 N

Dynamic viscosity of the oil (η) = 0.89 x 10^-3 Ns/m^2

To calculate the velocity gradient across the oil layer, we can use the equation:

τ = η * du/dy

where τ is the shear stress, η is the dynamic viscosity, and du/dy is the velocity gradient.

Since the force (F) applied on the plate is responsible for the shear stress, we can write:

τ = F / A

Substituting the given values, we have:

F / A = η * du/dy

Solving for du/dy, we get:

du/dy = (F / A) / η

Substituting the values, we have:

du/dy = (2.2 N) / (1.8 m^2 * 0.89 x 10^-3 Ns/m^2)

du/dy = 0.594 m/s/m

Therefore, the velocity gradient across the oil layer is 0.594 m/s/m.

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Answer:

B. The more potential energy it has.

Explanation:

I majored in Physics

Answer:

About 0.7767 seconds

Explanation:

\(\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s\)

Hope this helps!

Explanation:

dont know so sorryyyyyyyyy

The force required to stop the car of mass 3250 kg is -12195.12 N.

Force is the product of mass and acceleration.

To calculate the force required to stop the car, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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ANSWER

The mass of the student is 53.00kg

EXPLANATION

Given that;

The weight of the student = 520N

Follow the steps below to find the mass of the student

Step 1: Define weight

Weight s defined as the force acting on an object due to acceleration due to gravity

Step 2: Write the formula linking the mass and the acceleration due to gravity together

Where

W is the weight of the body

m is the mass of the body

g is the acceleration due to gravity

Step 3: Substitute the given data into the formula in step 2

Hence, the mass of the student is 53.00kg

Answer: 27,000 J :)

Explanation:

If the winch uses a 900 W motor and the motor is used for 30 s to pull a sailboat then work needed to do this work is 27000 J. hence option D is correct

Power is the rate of doing work. Power is also defined as work divided by time. i.e. Power = Work ÷ Time. Its SI unit is Watt denoted by letter W. Watt(W) means J/s or J.s-1. Something makes work in less time, it means it has more power. Work is Force times Displacement. Dimension of Power is [M¹ L² T⁻³]

Given,

Power P = 900W

Time = 30s

Work in joule =?

By using formula,

P = Work ÷ Time

Work = Power × time

W= 900 × 30

W= 27000 J

Hence Work of 27000J is needed to pull a sailboat to shore in 30s.

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The number of revolution = 31.83

The car starts from rest

The initial velocity, u = 0 m/s

The final velocity, v = 72 km/hr

v = 72 x 1000/3600

v = 20 m/s

The time, t = 5 secs

The radius of the wheel of the car, r = 0.25 m

Find the acceleration using the formula

v = u + at

20 = 0 + 5a

5a = 20

a = 20/5

a = 4 m/s

Find the distance using the formula

s = ut + 0.5at²

s = 0(5) + 0.5(4)(5²)

s = 0 + 50

s = 50m

The number of revolution is calculated using the formula below:

The number of revolution = 31.83

The East/West component of SuperPointParticleDog's force on the ball is -115.7 N. (Westward).

Isaac's force:

Magnitude: 255 N

Angle: 13 degrees North of East

x-component: 255*cos(13) = 245.1 N (Eastward)

y-component: 255*sin(13) = 58.1 N (Northward)

Newton's force:

Magnitude: 156 N

Angle: 34 degrees South of East

x-component: 156*cos(34) = 129.4 N (Westward)

y-component: 156*sin(34) = 86.5 N (Southward)

Now we can add the x- and y-components of the forces to find the net force:

Net force:

x-component: 245.1 N - 129.4 N = 115.7 N (Eastward)

y-component: 58.1 N - 86.5 N = -28.4 N (Southward)

The net force has an Eastward component of 115.7 N. Therefore, the East/West component of SuperPointParticleDog's force on the ball is -115.7 N. (Westward).

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Explanation:

Let

\(x_1\) = distance traveled while accelerating

\(x_2\) = distance traveled while decelerating

The distance traveled while accelerating is given by

\(x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2\)

\(\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2\)

\(\:\:\:\:\:= 1125\:\text{m}\)

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

\(v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}\)

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

\(v^2 = v_0^2 + 2ax_2\)

\(0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2\)

Solving for \(x_2,\) we get

\(x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}\)

\(\:\:\:\:\:= 4327\:\text{m}\)

Therefore, the total distance x is

\(x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}\)

\(\:\:\:\:= 5452\:\text{m}\)

The correct statement which best describes a magnetic object from among the options above is:

They have many spinning electrons oriented in the same direction that create magnetic fields.

The correct answer choice is option a.

When magnetic material spins its electrons in the same axis in the magnetic field, the force of current which passes through the same direction results in the attraction which exists between the magnetic object and magnetic substance.

So therefore, we can now confirm that when we consider the orientation of electrons of magnetic material in a direction similar to field, it causes attraction.

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Answer:

a)  B = 0.9375 T  -z, b)      B = 1.54 T

Explication

a) The magnetic force is

         F = q v x B

         f = q v B sin θ

bold indicates vectors

For direction let's use the right hand rule.

If the charge is positive

the flea in the direction of velocity, toward + x

the fingers extended in the direction of B

the palm the direction of the force + and

therefore the magnetic field goes in the direction of -z

           F = q v B

           2.25 10-16 = 1.6 10-19  1.50 103 B

           B = 2.25 10-16 / 2.4 10-16

           B = 0.9375 T  -z

b) now an electron

thumb speed direction, -z

opposite side of palm force + y

therefore the direction of the magnetic field is + x

            B = F / qv

            B = 8.5 10-16 / 1.16 10-19 4.75 103

             B = 1.54 T

Answer:

When the tension is  T =  16 N  

  The frictional force is  \(F_F = 22.2 \ N\)

When the tension is  T =  38 N  

    The frictional  force is  \(F_F = 11.1 \ N\)      

Explanation:

From the question we are told that

   The weight of the block is  \(w = 35 \ N\)

    The coefficient of static friction is \(\mu_s = 0.6\)

     The coefficient of kinetic friction is  \(\mu_k = 0.3\)

considering when the tension is  T =  16 N  

         \(W > T\) This implies that that block will not slide

Hence the frictional  force is  mathematically evaluated as

          \(F_F = \mu_s * W\)

=>       \(F_F = 0.6 * 37\)

=>        \(F_F = 22.2 \ N\)      

considering when the tension is  T =  38 N  

    \(W < T\) This implies that that block will start sliding

Hence the frictional  force is  mathematically evaluated as

          \(F_F = \mu_k * W\)

=>       \(F_F = 0.3 * 37\)

=>        \(F_F = 11.1 \ N\)      

Answer:

Explanation:

If a be grating element

a = 1 x 10⁻² / 11000

= .0909 x 10⁻⁵

= 909 x 10⁻⁹ m

for first order maxima , the condition is

a sinθ = λ where λ is wavelength

909 x 10⁻⁹ sin 49.67 = λ₁

λ₁ = 692.95  nm .

λ₂ = 909 x 10⁻⁹ sin 50.65

= 702.91 nm

λ₃ = 909 x 10⁻⁹ sin 52.06

= 716.88 nm

λ₄ = 909 x 10⁻⁹ sin 52.89

= 724.90 nm

692.95  nm ,  702.91 nm  , 716.88 nm , 724.90 nm .

Answer:

Water because the salt will just make it colder and the sugar won't do anything

Answer: In this scenario, we have two carts colliding with each other. One cart weighs 250 kg and is moving at a speed of 8 m/s, while the other cart weighs 100 kg and is initially at rest.

After the collision, the 250 kg cart continues moving forward, but at a slower speed of 3 m/s. We want to find out the speed at which the 100 kg cart moves after the collision.

To solve this, we use the principle that the total "push" or momentum before the collision should be the same as the total momentum after the collision.

Since the 100 kg cart is initially at rest, its momentum is zero. The momentum of the 250 kg cart before the collision is 250 kg * 8 m/s = 2000 kg·m/s.

After the collision, the momentum of the 250 kg cart becomes 250 kg * 3 m/s = 750 kg·m/s.

To find the momentum of the 100 kg cart after the collision, we subtract the momentum of the 250 kg cart after the collision from the total momentum before the collision: 2000 kg·m/s - 750 kg·m/s = 1250 kg·m/s.

Now, we divide this momentum by the mass of the 100 kg cart to find its velocity: 1250 kg·m/s / 100 kg = 12.5 m/s.

Therefore, the 100 kg cart moves at a velocity of 12.5 m/s after the collision, in the opposite direction of the 250 kg cart's motion.

Answer: 3.99•10⁴ J

Explanation:

The Heat Formula is: Q = m • C • ΔT

Your given C, m, T₀, and T

C = 0.385 J/g•K

m = 1.55 kg or 1,550 g

T₀ = 33.0 °C

T = 99.9 °C

(ΔT = T - T₀)

Your only remaining variable is Q, heat, so you can now plug in your values to solve.

Q = (1550)(0.385)(99.9 - 33.0)

(ΔT(°C) = ΔT(K) because the conversion is linear)

Q = 39,922.575 J

or

3.99•10⁴ J

Answer:

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  \(\tau = T \alpha\)

then,

⇒  \(\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R\)

⇒  \(\frac{1}{2}m \ a_t=m(g-a_t)\)

⇒  \(a_t=\frac{2g}{(\frac{m}{M} +2)}\)

On substituting the values, we get

⇒      \(=\frac{2\times 9.8}{\frac{0.7}{2} +2}\)

⇒      \(=8.34 \ m/s\)

hence,

⇒  \(T=mg+M(g-a_t)\)

On substituting the values, we get

⇒      \(=0.7\times 9.8+2(9.8-8.34)\)

⇒      \(=6.86+2(1.46)\)

⇒      \(=6.86+2.92\)

⇒      \(=9.8 \ N\)

Given data:

* The value of the force F_1 acting at 20 degree is,

* The value of the force F_2 is,

* The mass of the box is 8 kg.

Solution:

As box is not moving along the vertical direction, thus, the normal force acting on the body must balancing the y-comonents of the forces acting on the body,

The equation for the normal force is,

Here the negative sign is indicating the direction force towards downward, and the positive sign indicates the direction of force towards the upward direction,

Substituting the known values,

Thus, the normal force acting on the box is 57.88 N.