A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.

Answers

Answer 1

Answer:

4.9x10^-6T

Explanation:

See attached file

A Segment Of Wire Of Total Length 3.0 M Carries A 15-A Current And Is Formed Into A Semicircle. Determine

Related Questions

What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?

Answers

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field

Answers

Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?

Answers

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  [tex]\lambda = 622 nm[/tex]

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  [tex]D = 5 \ m[/tex]

    The order of the fringe is m  =  6

     The distance between the slit is  [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]

    The fringe distance is  [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]

Generally the for a dark fringe the fringe distance is  mathematically represented as

        [tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]

=>     [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]

substituting values

=>      [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]

=>     [tex]\lambda = 6.22 *10^{-7} \ m[/tex]

       [tex]\lambda = 622 nm[/tex]

if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c​

Answers

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/s

A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?

Answers

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N

Answers

The velocity and force are required.

The speed of the racket is 8.7 m/s

The required force is 471.43 N.

[tex]m_1[/tex] = Mass of racket = 1000 g

[tex]m_2[/tex] = Mass of ball = 60 g

[tex]u_1[/tex] = Initial velocity of racket = 12 m/s

[tex]u_2[/tex] = Initial velocity of ball = -15 m/s

[tex]v_1[/tex] = Final velocity of racket

[tex]v_2[/tex] = Final velocity of ball = 40 m/s

[tex]\Delta t[/tex] = Time = 7 ms

The equation of the momentum will be

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]

Force is given by

[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]

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Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.

Answers

Answer:

Zack should direct his throw outward and toward the back of the car.

Explanation:

As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.

The solution is throw 3.

I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

Which statement best applies Newton’s laws of motion?

The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.

When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.

The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.

Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

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A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t

Answers

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe

Answers

Answer:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.

Answers

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

Calculating the friction force:

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

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The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material

Answers

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down

Answers

Answer:

The  angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]

Explanation:

From the question we are told that

      The  angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]

       The  angular displacement is  [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]

From the first equation of motion we can define the movement of the record as

      [tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]

Given that the record started from rest [tex]w_o = 0[/tex]

So

       [tex]4.713^2 = 2 * \alpha * 25.14[/tex]

        [tex]\alpha = 0.4418 \ rad /s^2[/tex]

A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70

Answers

Answer:

The  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

      [tex]I = \frac{\epsilon}{ R }[/tex]

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration

Answers

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what

Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus

Answers

Answer:

2.55m

Explanation:

Using 1/do+1/di= 1/f

di= (1/f-1/do)^-1

( 1/0.0500-1/0.0510)^-1

= 2.55m

Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.

Answers

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

Answer:

D. Light is a form of energy

Explanation:

The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine

Answers

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron

Answers

Answer:

acting force is the answer

The direction of the magnetic force on the moving electron is upward.

The direction of the magnetic force on the electron can be determined by applying right hand rule.

This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.

Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.

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The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.

Answers

Answer:

27°

Explanation:

The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)

So theta = arcsin(0.45)

=27°

The angle between the wire and the magnetic field is 27°.

Calculation of the angle:

Since The magnetic force per meter on a wire is measured to be only 45 %

So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field

Therefore,

theta = arcsin(0.45)

=27°

Hence, The angle between the wire and the magnetic field is 27°.

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A ​46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this​ causeway?

Answers

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m

Answers

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?

Answers

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases

Answers

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answers

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration

Answers

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm

Answers

Answer:

Explanation:

The double slit interference phonemene is described for the case of constructive interference

          d sin θ= m λ                   (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

 

The double slit interference phonemene is described for the case of constructive interference

          d sin θ = m lam                    (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

we substitute

      sin θ = y / L

we substitute in equation 1

         d y / L = m λ

         λ = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 mm = 0.44 10⁻³ m

  y = 5.5 mm = 5.5 10⁻³ m

  L = 4.2m

  m = 1

let's calculate

        λ = 0.44  10⁻³ 5.5 10⁻³ / (4.2 1)

         

        λ = 5.76190 10-7 m

let's reduce to num

  lam = 5.56190 10-7 m (109 nm / 1m)

  lam = 556,190 nmtea

we substitute

      without tea = y / L

we substitute in equation 1

         d y / L = m lam

         lam = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 me = 0.44 10-3 m

  y = 5.5 mm = 5.5 10-3

  L = 4.2m

  m = 1

let's calculate

        lam = 0.44 10⁻³  5.5 10⁻³ / (4.2 1)

         

        lam = 5.76190 10⁻⁷ m

let's reduce to num

  lam = 5.56190 10⁻⁷ m (109 nm / 1m)

  lam = 556,190 nm

To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord

Answers

Answer:

The longer the cord, the lower the illumination

Explanation:

The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.

Long wires have more electrical resistance than shorter ones.

Let us consider this formula:

Resistance =[tex]\frac{\rho L}{A}[/tex]

From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.

In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?

Answers

Answer:

Explanation:

Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least  and kinetic energy is maximum .

In this way , there is conservation of mechanical energy .

Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.

Answers

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

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