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JulijaS [17]
3 years ago
13

A circular coil (800 turns, radius = 0.063 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is per

pendicular to the magnetic field. At t = 0.016 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.042 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.
Physics
1 answer:
bija089 [108]3 years ago
4 0

Answer:

0.000099T

Explanation:

For a circular coil the flux through a single turn changes by :

∆ = BAcos45° - BAcos90°

= BA cos 45°

During the interval of ∆t = 0.0165s

For N turn, Faraday's law gives the magnitude of emf as follows :

/E/= / -N BAcos45°/ ÷ ∆t

Since the loops are circular the area A of each loop is equal to πr square

B = /E/ ∆t ÷ Nπr square cos45°

= 0.042V × 0.0165/ 800 × 3.142 × 0.063^2 cos 45°

= 0.000693/7

= 0.000099T

= 9.9×10^-5T

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Answer:

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3 0
2 years ago
1. A ball falls from a height "h" from a tower. Which of the following statements is true?
Strike441 [17]

Answer:

the total kinetic and potential energy of the ball is constant (mechanical energy remains the same)

Explanation:

As the ball falls, kinetic energy is increased in direct relation with the decrease in potential energy

ΔKE + ΔPE = 0

4 0
3 years ago
A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____
OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

d_{o} = distance of the object = 4 cm

d_{i} = distance of the image = ?

f = focal length of the converging lens  = 8 cm

using the lens equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}

d_{i} = - 8 cm

4 0
2 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0
3 years ago
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
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