A 24-gauge copper wire has a diameter of 0.51 mm and is used to connect

A 24-gauge copper wire has a diameter of 0.51 mm and is used to connect a speaker to an amplifier. The speaker is located 7 m away from the amplifier. Part A) What is the minimum resistance of the connecting speaker wires at 20˚C? Part B) Compare the resistance of the wire to the resistance of the speaker (RSP = 8 Ω).

2 months ago

Solution 1

Guest Guest #2391067
2 months ago


a)  R_c = 1.18 ohms

b the resistance of copper wire is 14.7 % of resistance of speakers.



- 24-Gauge copper wire resistivity p = 1.72*10^-8

- The Length of the single wire L = 7 m

- The diameter of single copper wire d = 0.51 mm


A) What is the minimum resistance of the connecting speaker wires at 20˚C?

Part B) Compare the resistance of the wire to the resistance of the speaker (RSP = 8 Ω).


- We are given details of as single copper wire. However, it takes two copper wire to connect a speaker to the amplifier. After establishing that fact we can use the relation between the dimensions of the wire and the resistance as follows:

                                        R = p*L / A


R is the resistance of the wire in ohms

A is the cross sectional area of the wire

- Now for two wires the resistance would be twice:

                                        R_c = 2*p*L / A

- plug in the values:

                                        R_c = (2*1.72*10^-8 * 7 ) / (pi*(0.00051)^2 / 4)

- Evaluate:

                                        R_c = 1.18 ohms

- The wire resistance calculated can be compared with that of speaker by taking a ratio of the two:

                                        R_c / R_s

                                        1.18 / 8  * 100 = 14.7 %

- Hence, the resistance of copper wire is 14.7 % of resistance of speakers.

📚 Related Questions

Joe is camping in the mountains when a storm suddenly hits. Joe hears the thunder and sees the lightning. A lightning strike starts a forest fire; soon Joe can smell the smoke. Which areas of the brain will process the various sensory inputs Joe is receiving from the storm
Solution 1




Dealing with a storm is a stressful situation especially when the fire has started in the forest. These multiple signals are processed by hypothalamus that is located in the base of the brain and send the processed information to Pitutary gland that makes primary secretions to activate the renal medulla and a fight-or-flight response to stimuli is generated by rushing the body with energy induced by Adrenaline

A car is behind a truck going 25 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car can accelerate at 1.0 m/s^2. He gauges that he has to cover the 20 m length of truck, plus10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably also travelling at 25 m/s. He estimates that the car is about 400 m away, should he attempt the pass? Give details
Solution 1


No he should not attempt the pass


Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)

At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds

s = at^2/2

40 = 1t^2/2

t^2 = 80

t = \sqrt{80} = 8.94 s

Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is

8.94 * 25 = 223.6m

So the total distance traveled by the first car is 223.6 + 40 = 263.6m

The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is

8.94 * 25 = 223.6 m

So the total distance covered by both cars within this time frame

223.6 + 263.6 = 487.2m > 400 m

So no, he should not attempt the pass as we will not clear it in time.

A 2.4 m long rod of mass m1 = 14 kg, is supported on a knife edge at its midpoint. A ball of clay of mass m2 = 3.5 kg is dropped from rest from a height of h = 1.4 m and makes a perfectly inelastic collision with the rod 0.9 m from the point of support. Find the angular momentum of the rod and clay system about the point of support immediately after the inelastic collision.
Solution 1


It is known that;

            L_{i} = L_{f}

Now, we need to calculate the value of L, that is, angular momentum.


                   L= mvr

where,   m = mass

              v = velocity

              r = radius


               L = m(\sqrt{2gh}) \times r


              L = (2.2 \times (\sqrt{2 \times 9.81 \times 1.4}) \times 0.9)

                  = 10.37 Js

Thus, we can conclude that angular momentum of the rod and clay system about the point of support immediately after the inelastic collision is 10.37 Js.

A ball is thrown into the air and follows the parabolic path indicated by the dashed line. Use the arrows provided to indicate the approximate directions of the velocity and acceleration of the ball at each of the three locations. (Neglect air resistance.)
Solution 1

The ball is undergoing projectile motion such that the component of velocity and acceleration is towards the downward direction.

The problem is based on the projectile motion. Projectile motion is the motion possessed by any object in which it is only subjected to gravitational acceleration once released from a point.

  • In a projectile motion, the path described by an object under the action of forces is known as trajectory.
  • In the given problem, the ball is thrown in the air such that it follows the parabolic path. Then we can say that the ball has parabolic trajectory.
  • During its trajectory, the ball will experience only gravitational force due to Earth at downwards, so the velocity and acceleration of the ball will be downwards only. And there is no component of velocity and acceleration towards the horizontal direction.

Thus, we can conclude that the ball is undergoing projectile motion such that the component of velocity and acceleration is towards the downward direction.

Learn more about the projectile motion here:


Solution 2


Trajectory is the path of a projectile in air. Any tangential to the curve shows it velocity and the slope of the tangential line will be it acceleration.


The ambient temperature is 85.0°F and the humidity of the surrounding air is reported to be 56.0%. Using the Clausius-Clapeyron equation and the boiling point of water as 100.0°C at 760 torr, calculate the vapor pressure (in torr) of water in the air. Use 40.7 kJ/mol as the ∆Hvap of water.
Solution 1


The vapor pressure is 748.77 torr


Using Clausius-Clapeyron equation:

ln(\frac{P_2}{P_1}) = \frac{\delta H_v_a_p}{R}[\frac{1}{T_1}-\frac{1}{T_2}]


T₁ is the initial temperature = 85.0°F = 302.5 K

T₂ is the final temperature = 100 °C = 373 K

P₂ is the final pressure = 760 torr

P₁ is the initial pressure = vapor pressure = ?

R is gas constant = 8.314 J/K.mol

ΔHvap is the heat of vaporization of water = 40.7 kJ/mol

ln(\frac{P_2}{P_1}) = \frac{\delta H_v_a_p}{R}[\frac{1}{T_1}-\frac{1}{T_2}] = \frac{40.7}{8.314}[\frac{1}{302.5}-\frac{1}{373}]

ln(\frac{P_2}{P_1}) = 4.895(0.00331 - 0.00268) = 0.01489

\frac{P_2}{P_1} = e^{0.01489} = 1.015

P₁ = (760 torr)/(1.015) = 748.77 torr

Therefore, the vapor pressure is 748.77 torr

A water balloon is launched with an initial speed of 40 m/s at 60 degrees above the horizontal. A tall building is 40 m from the launch site. Neglect air resistance and use g 10 m/s2 8. How much time does it take for the water balloon to hit the building? a) 2.0 s b) 2.4 s c) 2.8 s d) 3.2 s e) 3.6s ,D 9. At what height above ground level does the water balloon hit the building? a) 49.3 m b) 45.3 m c) 41.3 m d) 37.3 m e) 33.3 m 10. How fast is the water balloon moving as it hits the building? a) 33.8 m/s b) 30.8 m/s c) 27.8 m/s d) 24.8 m/s e) 21.8 m/s
Solution 1


a) 2.0 s is the time after which the balloon hit the building

a) 49.3 m is the height at which the balloon hits the building

d) 24.8 m/s is the velocity of the balloon at the instant of hitting the building



initial speed of projection of balloon, u=40\ m.s^{-1}

angle of projection from the horizontal, \theta=60^{\circ}

distance of a tall building from the launch site, s=40\ m

Now we find the horizontal component of velocity :



u_x=20\ m.s^{-1}

Now the time taken to hit the building:

(since we don't have any acceleration in the horizontal direction )



t=2\ s

Now the vertical component of the velocity:



u_y=34.64\ m.s^{-1}

Time taken to reach the maximum height:

At max. height the final velocity (v=0)

v_y=u_y-g.t_m (-ve, since the direction of velocity is opposite to the gravity)

0=34.64-10\times t

t=3.464\ s

Therefore the the balloon hit the building while ascending in height.

Now the height reached in the time in which the balloon hits the building:

h=u_y.t-\frac{1}{2} g.t^2 (-ve, since the direction of velocity is opposite to the gravity)

h=34.64\times 2-0.5\times 10\times 2^2

h=49.28\ m

Speed of the balloon while it hits the building:

  • (The horizontal component of the velocity remains constant during the motion)

u_x=20\ m.s^{-1}

  • The vertical component of velocity at time t=2 sec:

v_y'=u_y-g.t (-ve, since the direction of velocity is opposite to the gravity)

v_y'=34.64-10\times 2

v_y'=14.64\ m.s^{-1}

Now the net resultant of the two components:



v=24.79\ m.s^{-1}

A particle of mass 89 g and charge 20 µC is released from rest when it is 47 cm from a second particle of charge −28 µC. Determine the magnitude of the initial acceleration of the 89 g particle. Answer in units of m/s 2 .
Solution 1


a = 256.36 m/s²



Charge of the 89 g particle = 20µC

distance from the second particle = 47 cm = 0.47 m

charge of the second particle = -28 µC

acceleration of 89 g particle = ?

Using Coulomb force formula

F = \dfrac{kQ_1Q_2}{r^2}

F = \dfrac{9\times 10^9\times 20\times 10^{-6}\times -28\times 10^{-6}}{0.47^2}

F = 22.82 N

For acceleration calculation

Using 2nd law of force

F = m a

22.82 = 0.089 a

a = 256.36 m/s²

Hence, the acceleration of the particle is equal to a = 256.36 m/s²

A dockworker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 10.5 m in a time of 4.80 s . (A) What is the mass of the block of ice? (B) If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ?
Solution 1


(a) mass of block ice=86.7 kg

(b) Distance=17.2179 m


Given data

For part (a)

First find  acceleration  then mass by ΣFx=ma

Let +x be the direction of force

ΣFx=79 N

x-x₀=10.5 m

t=4.80 s

Initial velocity V₀=0 m/s

x-x_{o}=v_{o}t+(1/2)at^{2}\\  10.5m=0+(0.5)a(4.80s)^{2}\\ 10.5m=11.52a\\a=10.5/11.52\\a=0.911m/s^{2}

Now ΣFx=ma


79N=m(0.911m/s^{2} )\\m=79/0.911\\m=86.7kg

For part (b)

First calculate the speed at the end period 4.50s of applied force

Then use the ending velocity as initial velocity in the second part of motion

After first 4.50 seconds

v=v_{o}+at\\ v=0+(0.911m/s^{2} )(4.50s)\\v=4.0995m/s\\

acceleration ax=0 m/s²

velocity v=constant



A U-tube manometer is used to measure the pressure at the stagnation point of a model in a wind tunnel. One side of the manometer goes to an orifice at the stagnation point; the other side is open to the atmosphere (fig. below). If there is a difference of 3 cm in the mercury levels in the two tubes, what is the pressure difference in N/m2 .
Solution 1


\Delta p=4000.4199\ Pa



  • difference in the column of mercury of u-tube manometer, \Delta h=3\ cm=0.03\ m

We know that the pressure due to height if a liquid column is given as:



p = pressure

h = height of the liquid

\rho= density of the liquid

  • We have the density of mercury as, \rho=13593\ kg.m^{-3}

Now the difference of pressure in the two tubes of the manometer:

\Delta p=\rho.g.\Delta h

\Delta p=13593\times9.81\times0.03

\Delta p=4000.4199\ Pa

(a) What is the energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×103V ? (b) Find the amount of stored charge.
Solution 1

(a) The energy stored is "405 J".

(b) Amount of stored charge is "0.09 C".


As we know, the stored P.E in capacitor given by:

U= \frac{QV}{2}




U= \frac{CV^2}{2}

By substituting the values, we get

      = \frac{10\times 10^{-6} F(9\times 10^3)^2}{2}

      = 405 \ J


By using the energy expression, we get

U = \frac{QV}{2}



      = \frac{2(405)}{9\times 10^3}

      = 0.09 \ C

Thus the above answers are correct.          

Learn more:


Solution 2


(a) U=405J

(b) Q=0.09C


(a) The potential energy stored in a capacitor is given by the expression:


Q is the stored charge and V the potential difference between capacitor plates. In a capacitor, we have:


Replacing this in the energy equation:


(b) Using the energy expression and solving for Q: