1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pishuonlain [190]
3 years ago
9

A long solenoid that has 1 080 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00

10-4 T at its center. What current is required in the windings for that to occur?
Physics
1 answer:
algol133 years ago
5 0

Answer:

Current, I = 0.073 A

Explanation:

It is given that,

Number of turns in a long solenoid is 1080

Length of the solenoid is 0.420 m

It produces a magnetic field of 10^{-4}\ T at its center.

We need to find the current is required in the winding for that to occur. The magnetic field at the center of the solenoid is given by :

B=\mu_0 NI

I is current

I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{10^{-4}}{4\pi \times 10^{-7}\times 1080}\\\\I=0.073\ A

You might be interested in
An infinite line of charge with charge density λ1 = -2 μc/cm is aligned with the y-axis as shown. 1) what is ex(p), the value of
Natasha2012 [34]

As we know by Gauss's law that

\int E.dA = \frac{q}{\epsilon_0}

so for line charge the gaussian surface is cylindrical in shape

so we will have

E(2\pi RL) = \frac{\lambda L}{\epsilon_0}

now by rearranging the terms

E = \frac{\lambda}{2\pi \epsilon_0 R}

so here we will have to find the x component of electric field so it is given by above equation

E_x = \frac{\lambda}{2\pi \epsilon_0 x}

here x = distance from the wire where we need to find electric field

4 0
3 years ago
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. par
Alenkasestr [34]
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
 We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
 The Coulombs Constant k = 9.0 x 10^9
 Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
  Mass of the Electron = 9.11 x 10^-31
  F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
  ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
 v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
 v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
 Since one orbit is the distance,
  one orbit = circumference = 2 x pi x r; distance s = v x t.
 v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
  t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
 Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
6 0
3 years ago
An electrostatic paint sprayer has a 0.17 m-diameter metal sphere at a potential of 25.0 kV that repels charged paint droplets o
Troyanec [42]

Answer:

q=0.236uC

Explanation:

From the question we are told that:

Diameter d=0.17m

Radius r=0.17/2=>0.085

Potential E=25.0kV

Generally the equation for Potential on spere is mathematically given by

E=\frac{1}{4 \pi e_0}*\frac{q}{r}

Therefore

q=\frac{25*10^3*0.085}{\frac{1}{4 \pi e_0}}

Where

\frac{1}{4 \pi e_0}=9*10^9

Therefore

q=\frac{25*10^3*0.085}{(9*10^9}}

q=0.236uC

4 0
3 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
4 years ago
A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the to
Alex Ar [27]

Answer:

The  change  in angular momentum is \Delta  L  = 0.0016 \ kgm^2/s

Explanation:

From the question we are told that

      The initial angular velocity of the spinning top is  w_1  =  16 \ rad/s

      The angular velocity after it slow down is  w_2 =  12 \ rad/s

      The time  for it to slow down is  t =  18 \ s

       The rotational inertia due to friction is  I =  0.0004 \ kg \cdot m^2

 Generally the change in the angular momentum is  mathematically represented as  

         \Delta  L  =  I  *(w_1 -w_2)

substituting values  

        \Delta  L  =  0.0004  *(16 -12)

       \Delta  L  = 0.0016 \ kgm^2/s

5 0
3 years ago
Other questions:
  • Questions 1 – 5 use the following word bank to place the correct letter in the space that correctly completes the sentence. A.PA
    15·1 answer
  • The conductive tissues of the upper leg can be modeled as a 40- cm-long, 12-cm-diameter cylinder of muscle and fat. The resistiv
    5·1 answer
  • a particle moves in an elliptical or circular motion. it is most likely a particle in which type of wave
    9·1 answer
  • Which is a characteristic of a step–down transformer?
    10·1 answer
  • Hooowwww dooo I doooo thiiiissss?!?!
    8·1 answer
  • Example of an unbalenced force
    15·1 answer
  • A 12m long gate that is 4 m wide is placed against a body of water at an angle of 60 degrees. The gate is being pulled by a rope
    8·1 answer
  • A certain cable of an elevator is designed to exert a force of 4.5x 104 N. If the maximum acceleration that a loaded car can wit
    12·1 answer
  • hollow place in rock made when an organism died and was buried and holes in the rock let air or water reach it and dissolve the
    13·2 answers
  • A batted ball is fair if it hits third base. *<br> True<br> O O<br> False
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!