If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

Answers

Answer 1
Assuming that the voltage is constant, and the resistance was doubled, since I=V/R, the current through the circuit will be halved. As P=IV, with the same voltage and halved current, the power dissipated by the circuit will be halved

Related Questions

A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed

Answers

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.

Answers

Answer:

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.

Explanation:

At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm

Answers

Answer:

f = 500 x 10^12Hz

Explanation:

E=hc/wavelength

E=hf

hc/wavelength =hf

c/wavelength =f

f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz

•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

Answers

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?

Answers

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

Calculation of the equivalent resistance:

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

Answers

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb

Answers

Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

What is work?

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field.

Answers

Answer:

Explanation:

area of the coil  A = .08 x .08 = 64 x 10⁻⁴ m ²

flux through the coil Φ = area of coil x no of turns x magnetic field

= 64 x 10⁻⁴ x 50 x B where B is magnetic field

emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through

Answers

The number of maxima of the standing wave pattern is two.

Maxima problem:

At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.

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A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

Answers

Answer:

Mb²/2

Explanation:

Pls see attached file

Need help understanding this. If anyone help, that would be greatly appreciated!

Answers

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

a) acceleration:

ā=v^2/r

ā=(15m/s)^2/27m

ā=225/27 m/s^2

ā=8.333 m/s^2

force:

F=mā. where the is equal to v^2/r

F=1000kg*8.3 m/s^2

F=8333.3 N

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?

Answers

Answer:

F= 175.5N

Explanation:

Given:

Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance

torque of 38N⋅m

angle of 120∘

Torque(τ): 38Nm

position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m

Angle: 120°

To find the Force, the torque equation will be required which is expressed below

τ = Frsinθ

We need to solve for F, if we rearrange the equation, we have the expression below

F= τ/rsinθ

Note: the torque is maximum when the angle is 90 degrees

But θ= 180-120=60

F= 38/0.25( sin(60) )

F= 175.5N

A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.

Answers

Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?

Answers

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

       

Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec

Answers

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Velocity

Given :

h=1/2t²+1/2t

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

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A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.

Answers

Answer:

A.21.3T

B.1.8x 10^6J/m^3

C.0.27x10^2J

D.6.6x10^-3H

Explanation:

Pls see attached file

Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?

Answers

Answer:

The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

Explanation:

We are told that the particle moves back and forth along a straight line with velocity v(t).

Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.

Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.

Answers

Answer:

Explanation:

Total mass m = 18 kg .

Spring are parallel to each other so total spring constant

= 4 x 24 = 96 N/cm = 9600 N/m

Time period of vibration

[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]

Putting the given  values

[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]

= .27 s .

An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner

Answers

Answer:

20.07ohms

Explanation:

Impedance is defined as the opposition to the flow of current through the elements of the circuit.

Impedance for R-L AC circuit is expressed as Z = √R²+XL²

R is the resistance

XL is the inductive reactance.

Given resistance of the air condition = 20 ohms

Inductive reactance XL = 1.68 ohms

Z = √20²+1.68²

Z = √400+2.8224

Z = √402.8224

Z = 20.07 ohms

Hence the impedance of the air conditioner is 20.07ohms

Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.

Answers

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10

Answers

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct

Answers

Answer:

B) the particle's momentum.

Explanation:

We know that

The centripetal force  on the particle when its moving in the radius R and velocity V

[tex]F_c=\dfrac{m\times V^2}{R}[/tex]

The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q

[tex]F_m=q\times V\times B[/tex]

At the equilibrium condition

[tex]F_m=F_c[/tex]

[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]

[tex]R=\dfrac{m\times V}{q\times B}[/tex]

Momentum = m V

Therefore we can say that the radius of curvature is directly proportional to the particle momentum.

B) the particle's momentum.

select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.

Answers

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.

Answers

Answer:

206.67N

Explanation:

The sum of force along both components x and y is expressed as;

[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]

The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

[tex]a_x = \frac{d^2 x }{dt^2}[/tex]

[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]

Similarly,

[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]

[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]

[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time

Answers

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

Induced EMF:

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity

Answers

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is

Answers

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago

Answers

Answer:

Your answer is( D) - Arago

1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.

Answers

Answer:

The answer is C

Explanation:

Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.

Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?

Answers

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

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