Which of the following phrases describes power?
A. How much energy is lost in friction
B. How much energy is lost in heat
OC. The rate energy is consumed
OD. The rate an object is moving

Answers

Answer 1

The rate energy is consumed this statement describes power.

Hence, Option C is correct answer.

How can we understand that this statement can describe power?

Power is related to energy by that it is the rate at which energy is transferred.

What is Power ?

It is a measure of the rate at which work is done.

According to the definition of power, Power is the amount of energy transferred or converted or consumed per unit time.

SI unit of power is watt.

By definition, 1 watt is equal to one joule of work done per second. So if P represents power in watts, E is the change in energy (number of joules) and t is the time taken in seconds then:

P=[tex]\frac{E}{t}[/tex]= [tex]\frac{1 Joule}{1 Second}[/tex]= 1 Watt.

Thus from the above conclusion we can say that, The rate energy is consumed describes power.

Hence, Option C is correct answer

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Related Questions

A sensor that generates voltages in the range of -2V..+2V for 0..400kg weight values Design a balance circuit with LCD display. ; fosc=4MHz​

Answers

39-72 may it not nah he 472=5

PLEASE HELP!!!!
Compared to driving at 20 mph, about how much longer will it take for you to stop at 60 mph?
about nine times as far
about six times as far
about twice as far

Answers

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.

Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].

Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].

Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.

Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].

Answer:

9 times

Explanation:

Friction typically __________ objects. A. Speeds up B. Slows down C. Doesn't affect D. Destroys

Answers

Answer:

B

Explanation:

friction opposes motion

Friction typically Slows down objects

Option "B"

How does friction affect speed?Friction and Speed

While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.

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find the volume of the tire with dimensions
● Tire diameter: 26”
● Tube width diameter: 2.125”

Answers

The volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

What is volume?

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

Volume of the tire

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

Radius of the tire = 0.5 x 26" = 13"

Volume of the tire = Area x thickness

Volume of the tire = πr² x h

where;

r is the radius of the tireh is the thickness of the tube

Volume of the tire = π(13)² x (2.125)

Volume of the tire =  1,128.2 cubic inch

Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

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A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 17.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.50 cm behind the glass shell.
a) Where is the coin located?
b) Determine the size of the image.

Answers

(a) The position of the coin placed next to the convex side of a thin spherical glass shell is 3.68 cm.

(b) The size of the image is 1.8.

Position of the coin

The position of the coin placed next to the convex side of a thin spherical glass shell is calculated as follows;

1/d = 1/f - 1/d'

where;

d' is the position of the coin's imaged position of the coinf is focal length

Focal length = r/2 = 17 cm / 2 = 8.5 cm

1/d = 1/8.5 - (-1/6.5)

1/d = 1/8.5 + 1/6.5

1/d = 0.1176 + 0.1538

1/d = 0.2714

d = 1/0.2714

d = 3.68 cm

Size of the image

Magnification, M = d'/d

M = (6.5)/(3.68)

M = 1.8

Thus, the position of the coin placed next to the convex side of a thin spherical glass shell is 3.68 cm.

The size of the image is 1.8.

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(a) Convert 45.0° to radians.
_______ rad

(b) Convert 13.0 rad to revolutions.
________ rev

(c) Convert 76.5 rpm to rad/s.
__________ rad/s

**Remember that there are 2 in one revolution, and 60 seconds in 1 minute. rad/s

Answers

(a) Converting 45.0° to radians is 0.785 radian.

(b) Converting 13.0 rad to revolutions is 2.1 rev.

(c) Converting 76.5 revolution per minute to rad/s is 8.02 rad/s.

Degrees to radian

To convert 45.0 degrees to radians, we use the following method;

180⁰ = π radians

45⁰ = ?

= (45 x π)/180

= 0.785 radian

Radian to revolutions

2π radians = 1 rev

13 radians = ?

= (13)/(2π)

= 2.1 rev

Revolution per minute to radians per second

= 76.6 rev/min x 2π rad/rev x 1 min/60 s

= 8.02 rad/s

Thus, converting 45.0° to radians is 0.785 radian.

Converting 13.0 rad to revolutions is 2.1 rev.

Converting 76.5 revolution per minute to rad/s is 8.02 rad/s.

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(a) Convert 46.0° to radians.
_______ rad

(b) Convert 17.0 rad to revolutions.
________ rev

(c) Convert 79.5 rpm to rad/s.
__________ rad/s

Answers

The answers are :

(a) 23π/90 radians ≈ 0.803 radians

(b) 17/2π revolutions ≈ 2.706 revolutions

(c) 79.5/60 rad/s = 13.25 rad/s

*The important formulas to remember*

Degrees to Radians

     [tex]\boxed {Radian = Degree \times \frac{\pi}{180^{o}}}[/tex]

    2. Radians to Revolutions

     [tex]\boxed {Revolution = \frac{Radian}{2\pi}}[/tex]

    3. rpm to rad/s

     [tex]\boxed {rad/s = \frac{rpm}{60}}[/tex]

Complete the following:

When light is incident through the focal point, and then strikes a mirror, ___

the light will reflect parallel to the principal axis
the light will reflect back through the focal point
the light will not reflect at all

Answers

Answer:

the light will reflect parallel to the principal axis

(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C

If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N

Direction?

- toward the left
- upward
-downward
- toward the right

Answers

The electric field strength at a point 1.00 cm to the left of the middle is  -2.0 x 10⁷ N/C.

The magnitude of the force is 94.4 N and direction of the force on it towards the right.

Electric field strength

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

Net electric field

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C

E = -2.0 x 10⁷ N/C

Force on the charge −4.72 µC

F = Eq

F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = 94.4 N

Thus, the direction of the force will be towards the right.

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A uniform linear charge of 3.0 nC/m is distributed along the y axis from y = -3 m to y=2m. Which of the following integrals is correct for the magnitude of the electric field at y = 4 m on the y axis?

Answers

[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex] is the integrals for the magnitude of the electric field.

A characteristic of electricity exists at every location in space when charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.

The formula for the electric field due to a point charge is:

[tex]E = \frac{kq}{r^2}[/tex]

Here q is the charge, r is the distance from the charge, and k = 8.99 × [tex]10^9 Nm^2/ C^2[/tex]

Imagine an infinitely - tiny portion of this charge distribution that has a charge [tex]dq[/tex]. This charge produces an electric field dE with magnitude:

                           [tex]\int\limits^ {} \, dE = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]

                               [tex]E = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]

Here, λ = 3.0 nC/m dq

           = λ.dy dE = ∫k.dq

From y = -3 m to y = 2 m, there is a 3.0 nC/m uniform linear charge distributed along the y axis.

The integrals for the magnitude of the electric field at y = 4 m on the y axis:

[tex]E = k \lambda \int\limits^2_{-3} {} \, \frac{dy}{( 4 - y)^2}[/tex]

[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex]

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What is an example of total internal reflection at work?

A.
A ray of light has the same intensity both entering and exiting a fiber optic cable.
B.
A ray of light entering a glass cube gets refracted.
C.
A ray of light in air hits a shiny surface and bounces off.
D.
A ray of light entering a ruby gets refracted.

Answers

[tex]\huge\underline{\underline{\boxed{{ \mathbb{SOLUTION:}}}}}[/tex]

[tex]\leadsto[/tex] Total internal reflection occurs when a ray of light is traveling through a medium and, when that medium changes, it does not refract into the second medium but instead it reflects back into the first medium.

By Snell's law, we know that

[tex]\longrightarrow \sf{n_1 sin \emptyset_1 = n_2 sin \emptyset_2}[/tex]

If this happens, then [tex]\sf{n_2 \angle n_1}[/tex], and the ray does not lose intensity due to refraction.

[tex]\huge\underline{\underline{\boxed{\mathbb{ANSWER:}}}}[/tex]

[tex]\large \bm{ A.}[/tex] A ray of light has the same intensity both entering and exiting a fiber optic cable.

If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)

a) How far and at what angle is the Aster's final position from her initial position?

b) In what direction would she has to head to return to her initial position

Answers

Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position

What is Displacement ?

Displacement is the distance travelled in a specific direction. It is a vector quantity.

Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

a) Let P be the Aster's final position from her initial position?

We can use bearing by using Cosine formula to solve this question.

P² = 70² + 82² - 2 × 70 × 82 cos 73

P² = 4900 + 6724 - 11480 cos 73

P² = 11624 - 3356.43

P² = 8267.57

P = √8267.57

P = 90.9 m

P = 90.9 - 28

P = 62.9 m

We can get the angle by using Sine rule

82/ sin Ф = 90.9 / sin 73

sin Ф = 0.8627

Ф = [tex]Sin^{-1}[/tex] (0.8627)

Ф = 59.6°

Ф = 60°

b)  She will head north west direction to return to her initial position

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Please help with this question, step by step

Thank you!

Answers

From the calculation, the force constant is 192 N. Also, friction would decrease the extension.

What is the force constant?

We know that the force constant can be obtained by the use of the relation;

F = Ke

F = applied force

K = force constant

e = extension

We know from Hooks law that the force applied is directly proportional to the extension.

We can write;

F = mgcosθ

F = 43 Kg * 9.8 m/s^2 * sin31°

F = 217 N

K = 217 N/1.13 m

K = 192 N/m

If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.

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At the very end of 500 DAYS OF SUMMER, Tom (Joseph Gordon-Levitt) is so heartbroken by
Summer's rejection, that he vows to avoid any more romantic entanglements. As he observes "There
are no miracles. There's no such thing as fate. Nothing's meant to be."
True
False

Answers

He observes that "There are no miracles. There's no such thing as fate. Nothing's meant to be." this is a false statement.

What is the Definition of miracle?

The term miracle is known to be something that is said to be an extraordinary event and it is one that often manifest in a kind of a divine intervention in the lives and affairs of human such as healing miracles, miracle jobs, miraculous meeting, etc.

Note that bad things that happens to people often makes them not to believe in the extraordinary but it does exist.

Hence, He observes that "There are no miracles. There's no such thing as fate. Nothing's meant to be." this is a false statement.

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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.52 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

________V/m

Answers

The electric filed in V/m is 1.58 * 10^6 V/m

What is the electric field?

We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.

Thus;

E = V/d

E = electric field

V = voltage

d = distance of separation

E = 24 * 10^3 V/1.52 * 10^-2 m

E = 1.58 * 10^6 V/m

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It is useful to use a _________ to demonstrate how an electric circuit works

Answers

It is useful to use a model in order to demonstrate how an electric circuit works.

What is a model?

In science, a model is a graphical representation that my result useful to predict the behavior of the components of a given system that works together to produce a particular outcome.

For example, an electrical circuit can be modeled by taking into account resistance, voltage, and intensity as parameters of functioning.

The electrical circuits are graphically represented by modeling the movement or flow of negatively charged electrons (e-) from the negative pole to the positive pole of a closed electric circuit.

In conclusion, it is useful to use a model (ie., a scientific model) to demonstrate how an electric circuit works.

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. An aeroplane takes 30 min to travel from Biratnagar to Kathmandu. If the air distance between
the Biratnagar and Kathmandu is 450 km. Find the average velocity of the plane.

Answers

Answer:

average velocity = [tex]\bf 900 \space\ km/h[/tex]

Explanation:

We can find the average velocity using the following equation:

[tex]\boxed{average \space\ velocity = \frac{total \space\ distance \space\ travelled}{ \space\ time \space\ taken}}[/tex] .

In this case:

• total distance travelled = 450 km

• time taken = 30 min = 0.5 h

Substituting these values into the equation:

[tex]average \space\ velocity = \frac{450 \space\ km }{ 0.5 \space\ h}[/tex]

                          ⇒ [tex]\bf 900 \space\ km/h[/tex]

method to separate the .
sand+iron fillings​

Answers

Answer:

by using a magnetic field

Use a magnet to attract the iron fillings from the sand!

Hope this helps!

What is the equation for the enthalpy of vaporization?

A. Q = m*Lv
B. Q = m/Lv
C. Q = 1/m*Lv
D. Q = Lv/m

Answers

Answer:

A: Q = m*Lv

Explanation:

The equation formula for the enthalpy of vaporization is;

Q = m*ΔHv

Where;

q = heat energy

m = mass

ΔHv = heat of vaporization

2. Two particles P and Q are shot vertically up. T
Particle P is first up with a velocity 40ms-¹. After
4s, particle Q is also shot up. Find 01 -
(a) where the two particles meet, if at the point of
meeting, P has velocity of 15ms™¹.
(b) the velocity with which Q is shot. (g = 10ms-²)
(000)(9)

Answers

Free Fall under gravity and time of meeting is used to find out the required answer.

∴ Let us assume, the upward direction to be positive and and downward direction to be negative.

Given, Velocity of P, = 40m/s

Distance travelled by P,

∴Using First Equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s  ; which is the time taken by P to rise up

Now, Maximum Height(s) reached by particle P,

∴Using second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) Particle P is falling when Q is shot up after 4s from the initial time

∴ Using third equation of motion using Free Fall under Gravity,

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Hence, Particle P and Q meet at a distance (H₂) from ground,

∴ Height, H₂ = s - H

⇒ H₂ = 80 - 11.25 = 68.75m

Hence, Particle P and Q meet at a distance of 68.75m from ground.

∴ Using First equation of motion for Particle P using the Time of Meeting,

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s  ; which is equal to the fall time of P and Rise time of Q

b) For particle Q

∴ Using second equation of motion,

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Hence, the Velocity with which Particle Q was shot is 15m/s in the upward direction.

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What is the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean?
a. It is slightly less than 9.81 m/s2.
b. It is very large due to the high pressure.
c. It is zero.
d. It is significantly less than 9.81 m/s2.

Answers

The gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure. That is option B

What is gravitational acceleration?

The gravitational acceleration is the acceleration of an object that is under free fall. The acceleration due to gravity is 9.81 m/s2. above the sea level.

One of the factors that affect gravitational acceleration is depth. This is because as the depth increases so will pressure increase.

Increase in pressure leads to increase in gravitational acceleration.

Therefore, the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure.

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A centrifuge in a medical laboratory rotates at an angular speed of 3,750 rev/min. When switched off, it rotates through 48.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s²) of the centrifuge.


______rad/s²

Answers

The constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².

Constant angular acceleration of the centrifuge

The constant angular acceleration of the centrifuge is calculated as follows;

ωf² = ωi² - 2αθ

where;

ωf is the final angular velocity ωi is initial angular velocityθ is angular displacementα is angular acceleration

When the centrifuge is switched off, the final angular velocity = 0

Initial angular velocity: ωi = 3,750 rev/min x 2π rad/rev x 1 min/60 s = 392.7 rad/s

angular displacement: θ = 48 rev = 48 rev x 2π rad/rev = 301.6 rad

0 = ωi² - 2αθ

2αθ = ωi²

α = ωi²/2θ

α = (392.7²) / (2 x 301.6)

α = 255.66 rad/s²

Thus, the constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².

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Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to position F. Also assume that the roller coaster has minimal kinetic energy at point A. Create three different models below to describe the kinetic energy, the potential energy, and the total energy of the roller coaster at point E. Examples of models include, but are not limited to: pie charts, bar graphs, equations, textual descriptions, and analogies.

Answers

At point E

the kinetic energy of the rollercoaster is small compared to the potential energythe potential energy is greater than the kinetic energythe total energy is a mixture of potential and kinetic energy

What is the energy of the roller coaster at point E?

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

the kinetic energy of the rollercoaster is small compared to the potential energythe potential energy is greater than the kinetic energythe total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

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2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across a smooth horizontal surface by a force F evaluated the magnitude of F and what force the blocks exert on each other.​

Answers

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

what is buoyant force​

Answers

Answer:

When an object is immersed in water. it is pulled downwards due to gravitational pull of earth. Water exerts upward force on the object. This makes object rise up. This upward force is called buoyancy or upthrust.

Nina is doing a workout where she runs at 50% speed for 3 minutes, 75% speed for 2 minutes, and 100% speed for 30 seconds. This is an example of
a. Specificity training
b. Overload training
c. Circuit training
d. Interval training

Answers

D. Interval training
d. Interval training
Not sure if I’m right

What is meant by constructive interference?
A. Waves combine and their amplitude decreases
B. Waves do not combine and their amplitude decreases
C. Waves do not combine and their amplitude increases
D. Waves combine and their amplitude increases

Answers

Answer:

D. Waves combine and their amplitude increases

Explanation:

amplitude is the height of the wave in a graph

constructive interference happens when two waves are lined up perfectly

example when two speakers playing same music while facing each other, music will appear louder

Destructive interference happens when the peaks match the valleys and they cancel perfectly

like with noise cancelling headphones

during constructive interference, waves collide in phase

in phase means add together making the height of the wave in a graph bigger (amplitude increases)

Example: When we see two speakers right next to each other, we can experience constructive interference when the distance from each speaker to the observer is the same

two waves destructively interfere, they must have the same amplitude in opposite directions

quizlet

flexbooksck12org

byjus

chegg

Complete the missing information from the circuit

Answers

From the first circuit (see attached), it is correct to state that I₄ = 40mA. See the explanation below.

What is a circuit?

A path that is designed to transport electric current is that is referred to as a circuit.

How do we compute for the missing details?

I = 200mA

= 20mA + 140mA + I₄

I₄ = 200 - 160

I₄ = 40mA

From the above, we can rightly posit:

R₄ = 3/(40x 10⁻³)

= 75 Ω

Hence

V₆ =  V₄ - V₅

= 3v - 2v

= 1v

R₆ therefore, =

V₆/I₆

= 1/ (40 x 10 ⁻³)

= 14.28 Ω

I₃ = I₄ + I₆

= 40 + 140

= 180mA

R₃ = 9/(180 x 10⁻³)

= 50Ω

R₂ = 12/20 x (10⁻³)

= 60 Ω

Then,

Vbattery =

200 x 10⁻³ x90

= 18 volts

The above indices is used to complete the image. See Circuit Image II.

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Show that time period of revolution an electron beam in uniform magnetic field is independent of velocity and radius of path

Answers

Since T = 2πm/Be the period of revolution of the electron is independent of velocity and radius of path

How to show that time period of revolution an electron beam in uniform magnetic field is independent of velocity and radius of path?

For an electron moving in a uniform magnetic field, the force on the electron is given by

F = Bev where

B = magnetic field, e = electron charge and v = speed of electron

Also, this magnetic force equals the centripetal force on the electron, F'

F' = mv²/r where

m = mass of electron, v = speed of electron and r = radius of path

Since both forces are equal,

F = F'

Bev = mv²/r

Be = mv/r

We know that angular speed, ω = v/r. So,

Be = mω

Also, angular speed, ω = 2π/T where T = period of revolution of electron

So, Be = m2π/T

Making T subject of the formula, we have

T = 2πm/Be

so, since T = 2πm/Be the period of revolution of the electron is independent of velocity and radius of path

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a bus come to rest if the acceleration is 4m/s² find the distance travelled​

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Answer:

One needs to know the initial speed of the bus to find distance traveled.

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