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Schach [20]
3 years ago
14

A quadraphonic car stereo operates on electric current provided by the car's 12V battery and is connected in series. Each channe

l of the stereo, which feeds the electrical signal to one of the stereo's four speakers, has a resistance of 4.1 oums. How much current is in the circuit of each stereo channel?
Physics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

2.93 A

Explanation:

From Ohm's law V = IR where V = voltage = 12 V, I = current = ? and R = resistance = 4.1 Ω.

So, the current in each quadraphonic stereo circuit is I = V/R = 12 V/4.1 Ω = 2.93 A

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An open train car rolls along a track while it is being filled with sand. There is negligible friction. As a result of
Angelina_Jolie [31]

Answer:

Explanation:

<h3>that`s a the train car, that you asked the meaning, of that if the train car rolls it`s doing it`s speed, and it`s not ganna fall off the the trail of the train, car.</h3>
3 0
2 years ago
A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resi
finlep [7]

<u>Answer</u>

3 Ohms

<u>Explanation</u>

when the resistors are in series, the resistance in the circuit increases. For example, if two resistors, R1 and R2 are in series, the combined resistance is R1+R2.

When connected in parallel, the total resistance is the reciprocal of (1/R1 + 1/R2)

In this case the resistors are in parallel.

Total resistance = (1/12 + 1/4)⁻¹

= (1/3)⁻¹

= 3 Ohms

7 0
3 years ago
Read 2 more answers
what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el
Evgen [1.6K]

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

5 0
1 year ago
During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 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
3 years ago
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