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lesantik [10]
3 years ago
14

2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. Wh

at direction would you expect the induced magnetic field to be in?4. To decrease the magnitude of current induced in the inductor using a bar magnet, you could (select all that apply):a. Decrease the strength of the magnet.b. Decrease the velocity of the magnet going into the solenoid.c. Decrease the number of coils in the solenoid.d. Increase the cross sectional area of the solenoid.6. List the TWO values you need to record to measure a voltage on the oscilloscope
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

2) deflection must be towards the negative side of the voltage.

4) the correct statements are: b and c

Explanation:

2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by

           E = - N d \phi_{B} / dt

where \phi_{B} = B. A

The bold are vectors

Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.

4) From the Faraday equation we can see that the inductive electromotive force depends

* The magnitude of B that changes over time

* The area of ​​the loop that changes over time

* The angle between B and the area that changes over time

* A combination of the above

With this analysis we will review the different alternatives given

a) False. It takes a temporary change and an absolute value of B

b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases

c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller

d) False. A temporary change of area is needed, in addition to increasing the area the current increases

We can see that the correct statements are: b and c

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Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.
4 0
3 years ago
Please helps me with this question..
Alenkasestr [34]
1) S.I. Unit for electric current = "Ampere"

2) S.I. Unit for resistance = "Ohm"

3) S.I. Unit for potential difference = "Volt"

Hope this helps!
8 0
3 years ago
3. An object of mass 90 kg travels down a slide.
Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

E_{k}=\frac{1}{2} *m*v^{2}  \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]

4 0
3 years ago
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest
shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0
4 years ago
The most dangerous thing for a motorcyclist passing parked cars is
jeka94
The main danger is vehicles making u-turns or pulling out without signalling.
6 0
4 years ago
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