Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Answers

Answer 1

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1


Related Questions

what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l

Answers

Answer:

1.94cm³/s

Explanation:

1L = 1000cm³

Ihr = 3600s

So

Using

Average flow rate

Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L

= 1.94cm³/s

A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend

Answers

Answer:

Explanation:

The e.m.f induced in the coil depend on the following :

(a) No. of turns in the coil

(b) Cross-sectional Area of the coil

(c) Magnitude of Magnetic field

(d) Angular velocity of the coil

The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.

Answers

Answer:

A) it transforms a small force acting over a large distance into a large force acting over a small distance.

Explanation:

The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.

Pressure transmitted P = F/A

where F is the force applied

and A is the area over which the force is applied.

This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.

F/A = f/a

where the left side of the equation is for the output, and the right side is for the input.

The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that

volume V = (area A) x (distance d)

this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.

From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______

Answers

Answer:

-z axis

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change

Answers

B. A chemical change

Explanation:

I'm guessing ?

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?

Answers

Answer:

Explanation:

According to Equations of Projectile motion :

[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

[tex]Horizontal Range = vcosx * t[/tex]

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.

Answers

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.

Answers

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that

Answers

Answer:

A) the moment of inertia of the system decreases and the angular speed increases.

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

[tex]I_{1} w_{1} = I_{2} w_{2}[/tex]    ....1

where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.

and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

[tex]I = mr^{2}[/tex]    ....2

where [tex]m[/tex] is the mass of the rotating body,

and [tex]r[/tex] is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled

Answers

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

           

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

         

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?

Answers

Answer:

8.78 Amps

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A

Hence the current when the unit/heater is operating is 8.78 Amp

Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion​

Answers

Last point, Newton’s first law of motion is the correct answer.

Hope this helps ya

if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?

Answers

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)

Answers

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz

A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s

Answers

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

   

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

Calculation of Wavelength

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

Finf more informmation about Wavelength here:

https://brainly.com/question/6352445

A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?​

Answers

Is there any other type of information?

Explanation:

Using Equations of Motion :

[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .

Answers

Answer:

[tex]\theta=10.20^{\circ}[/tex]  

[tex]\Delta l=0.10 ft[/tex]    

Explanation:

First of all, we analyze the system of blocks before starting to move.

[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]  

[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]  

[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]  

[tex]tan(\theta)=0.18[/tex]  

[tex]\theta=arctan(0.18)[/tex]  

[tex]\theta=10.20^{\circ}[/tex]  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]

Where:

[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]

[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]

[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]

[tex]\Delta l=0.10 ft[/tex]    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

(a) The inclined angle for which both blocks begin to slide is 10.3⁰.

(b) The compression of the spring is 0.22 ft.

The given parameters;

mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ft

The normal force on block A and B:

[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]

The frictional force on block A and B:

[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]

The net force on the blocks when they starts sliding;

[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]

The change in the energy of the blocks is the work done in compressing the spring;

[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]

Learn more here:https://brainly.com/question/16892315

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?

Answers

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings

Answers

Answer:

The AC generator has one unbroken slip ring

Explanation:

In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.

An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.

We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.

It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.

You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?

Answers

Answer:

3.067MHz

Explanation:

The formula for calculating the voltage across an inductor is expressed as

[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor [tex]V_l[/tex] = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]

Hence, the frequency required is 3.067MHz

An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.

Answers

Maybe it is around 300

Answer:

433

Explanation:

The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.1 mm. Using light with a wavelength of 578 nm, how far could you be from this tile and still resolve these holes

Answers

Answer:

8.65x10^3m

Explanation:

See attached file

A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.

Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.

E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.

Answers

Answer:

A. 990v/m

B.330x10^-8T

C.2.19x10^-6J/m³

D.1.45x10^-11J

Explanation:

See attached file

PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.

Answers

Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.

Answers

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?

Answers

Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?

The direction for each field vector is perpendicular to equipotential lines.

Take a snapshot of the simulation showing equipotential lines and paste to a word document.

Answers

....................

A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?

Answers

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?

Answers

Answer:

50 cm long

When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.

An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s

Answers

Answer:

The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

From the question we are told that

     The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]

     The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]

      The  speed is  [tex]v = 18.8 \ m/s[/tex]

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as

       [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]

Substituting values

      [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]

       [tex]Q = 2.177 *10^{-9} \ C[/tex]

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