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24-gauge copper wire has a diameter of 0.51 mm. The speaker is located exactly 4.27 m away from the amplifier. What is the minimum resistance of the connecting speaker wire at 20°C? Hint: How many wires are required to connect a speaker!Compare the resistance of the wire to the resistance of the speaker (Rsp = 8 capital omega)

Answers

Answer 1

Answer:

Answer:

R = 8.94 10⁻² Ω/m, R_sp / R_total = 44.8

Explanation:

The resistance of a metal cable is

R = ρ L / A

The area of a circle is

A = π R²

The resistivity of copper is

ρ = 1.71 10⁻⁸ ohm / m

Let's calculate

R = 1.71 10⁻⁸ 4.27 / (π (0.51 10⁻³)²)

R = 8.94 10⁻² Ω/m

Each bugle needs two wire, phase and ground

The total wire resistance is

R_total = 2 R

R_total = 17.87 10⁻² Ω

Let's look for the relationship between the resistance of the bugle and the wire

R_sp / R_total = 8 / 17.87 10⁻²

R_sp / R_total = 44.8

Answer 2

Answer:

Final answer:

The resistance of the speaker wire can be calculated using the formula for the resistance of a wire, taking into account the resistivity of copper, the length and thickness of the wire, and whether a single or pair of wires is used.

Explanation:

The question is asking you to find the minimum resistance of a copper wire given its diameter and length, plus the resistance of the speaker it's connected to. Resistance of a wire is calculated using the formula R=ρL/A, where R is the resistance, ρ (rho) is the resistivity of the material (in this case, copper), L is the length of the wire, and A is the cross-sectional area of the wire.

First, you need to find the area of the 0.51 mm diameter wire. The area (A) of a wire is given by the formula π(d/2)^2 where d is the diameter of the wire. After calculating the area, use the formula R=ρL/A to calculate the resistance. For copper wire at 20°C, ρ is approximately 1.68 × 10^-8 Ω·m. Substituting these values into the formula will give you the resistance of the wire in ohms.

Note: you may need to consider whether you have just a single wire or a pair, since two wires are typically required to connect a speaker. If a pair is used, each wire will carry half the current, which affects the total resistance.

Learn more about Electric Resistance here:

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ou are writing a short adventure story for your English class. In your story, two submarines need to arrive at a place in the middle of the Atlantic Ocean at the same time. They start out at the same time from positions equally distant from the rendezvouspoint. They travel at different speeds, but both go in a straight line. The first submarine travels at an average speed of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the story’s plot, the second submarine is required to travel at a constant velocity, so the captain needs to determine the magnitude of that velocity. What is that velocity?

Answers

Answer:

The constant speed of second submarine is 31.16 km/hr

Explanation:

Given that

v₁=20 km/hr ,d₁= 500 Km

v₂=40 km/hr ,d₂=500 km

v₃=30 km/hr, d₃=500 km

v₄=50 km/hr ,d₄=500 km

We know that

Displacement = Velocity x Time

d = v t

Total displacement = Average velocity x Total time

$d_1+d_2+d_3+d_4=V_(avg)\left((d_1)/(v_1)+(d_2)/(v_2)+(d_3)/(v_3)+(d_4)/(v_4)\right)$

Now by putting the values

$2000=V_(avg)\left((500)/(20)+(500)/(40)+(500)/(30)+(500)/(50)\right)$

$V_(avg)=31.16\ km/hr$

So the constant speed of second submarine will be the average speed of first submarine because they have to meet at the same time.

The constant speed of second submarine is 31.16 km/hr

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answers

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=(1)/(2)kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_(i)=E_(f)$

$U_(i)+U'_(i)=U_(f)+U'_(f)$

$(1)/(2)kx^2+0=0+mgh$

$h=(kx^2)/(2mg)$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=(5365*(0.097)^2)/(2*0.221*9.8)$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

1. What is the frequency of light waves with wavelength of 5 x 10-⁷ m?

Answers

Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

Definition of wavelength

First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Definition of frequency

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Definition of propagation speed

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.

Frequency of light waves with wavelength of 5×10⁻⁷ m

In this case, you know:

v= 3×10⁸ m/s
f= ?
λ= 5×10⁻⁷ m

Replacing in the definition of propagation speed:

3×10⁸ m/s = f× 5×10⁻⁷ m

Solving:

3×10⁸ m/s ÷ 5×10⁻⁷ m= f

f= 6×10¹⁴ Hz

In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

Learn more about wavelength, frecuency and propagation speed:

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Answer:

Speed of light =m/s

wavelength = m

frequency = ?

we have

Speed = frequency × wavelength

$3* 10^8$ = frequency × $5 * 10^(-7)$

Frequency = $(3*10^8)/(5*10^(-7))=6*10^(14)$ hz

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answers

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

$v^2-u^2=2as$

$18^2-0=2* a* 125$

$a=1.296 m/s^2$

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m

$v^2-u^2=2as$

$v^2=2* 1.296* 200$

$v=√(518.4)=22.76 m/s$

Imagine that you drop an object of 10 kg, how much will be the acceleration andhow much force causes the acceleration?

Answers

If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.

If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?