PHYSICS COLLEGE

An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer 1

Answer:

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

Related Questions

Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diffraction pattern?A. fraunhofer B. fresnelC. far-fieldD. single slit
A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?
How long would it take a 500. W electric motor to do 15010 J of work?
A 1.7 kg model airplane is flying north at 12.5 m/s initially, and 25 seconds later is observed heading 30 degrees west of north at 25 m/s. What is the magnitude of the average net force on the airplane during this time interval?
A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.Determine the following:a. The total amount of energy transfer by work (kJ)b. The total amount of energy transfer by heat (kJ)

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).

Answers

Answer:

k = 15.62 MN/m

Explanation:

Given:-

- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.

Constraints:-

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

Find:-

What is the minimum required stiffness of each of the two springs as per the constraints given.

Solution:-

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

$y ( t ) = Y_o*sin ( w*t )$

Where,

Yo : The amplitude of excitation = 3 mm

w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

$y ( t ) = 3*sin ( 36\pi *t )$

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

$m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)$

Where,

m: The combined mass of the rigid body ( wheel + grinding wheel body) c : The viscous damping coefficient

k_eq: The equivalent spring stiffness of the system ( parallel )

x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

$w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) = (1800)/(2√(k_e_q*428.135) ) = (43.49628)/(√(k_e_q) )$

Where,

w_n: The natural frequency

p = ζ = damping ratio = c / cc , damping constant/critical constant

- The Equation of motion becomes:

$(1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)$

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

$X_s_s = X_o*sin ( wt + \alpha )$

Where,

X_o : The amplitude of the steady-state vibration.

α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

$X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }$

Where,

r = Frequency ratio = $(w)/(w_n) = \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135) )/(√(k_e_q) )$

- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

$((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135) )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)\n\n$

$((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201 )/(k_e_q) )]^2 + (41442858448.85813)/(k_e_q^2 ) )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2 )/(k_e_q^2) ] + (41442858448.85813)/(k_e_q^2 ) )}\n$

$((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0$

- Solve the inequality ( quadratic ):

$k1_e_q \geq 7811740.790197058 (N)/(m) \n\nk2_e_q \leq 4224034.972855095 (N)/(m)$

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

$k_e_q = (k^2)/(2k) = (k)/(2)$

- Therefore,

$k1 \geq 7811740.790197058*2 = 15.62 (MN)/(m) \n\nk2 \leq 4224034.972855095*2 = 8.448 (MN)/(m)$

- The minimum stiffness of spring is minimum of the two values:

k = 15.62 MN/m

Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?A. 165 m/s
B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s

Answers

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

A certain lightning bolt moves 40 C of charge. How many fundamental units of charge |qe| is this?

Answers

Answer to A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge | qe | is this? . ... charge, N is the total number of electron or protons that constitute total charge Q.

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

Learn more here:brainly.com/question/11788630

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration

Answers

Answer:

The answer is below

Explanation:

Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate

The sum of moment is given as:

$50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)$