Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
where would you expect to find vesicles of neurotransmitters
A. Synaptic gap
B. postsynaptic dendrites
C. Channels in the postsynaptic
D. Presynaptic terminal button
Answer:
D. Presynaptic terminal button
explanation:
Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron
hope this helped!
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
An aluminium pot whose thermal conductivity is 237 W/m.K has a flat, circular bottom
with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in
the pot through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pot
is at 105 °C, determine the temperature at the outer surface of the bottom of the pot
Answer:
T₁ = 378.33 k = 105.33°C
Explanation:
From Fourier's Law of heat conduction, we know that:
Q = - KAΔT/t
where,
Q = Heat Transfer Rate = 1400 W
K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k
A =Surface Area through which heat transfer is taking place=circular bottom
A = π(radius)² = π(0.15 m)² = 0.0707 m²
ΔT = Difference in Temperature of both sides of surface = T₂ - T₁
T₁ = Temperature of outer surface = ?
T₂ = Temperature of inner surface = 105°C + 273 = 378 k
ΔT = 388 k - T₁
t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m
Therefore,
1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m
(1400 W)/(4188.14 W/k) = - (378 k - T₁)
T₁ = 0.33 k + 378 k
T₁ = 378.33 k = 105.33°C
Convert from scientific notation to standard form
9.512 x 10-8
A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.A) What is the magnitude of the force of the car on the truck?B) What is the magnitude of the force of the truck on the car?
Answer:The answer is 3000 N.
Force (F) is the multiplication of mass (m) and acceleration (a).
F = m · a
It is given:
mc = 1000 kg
mt = 2000 kg
total force: F = 4500 N
total mass: m = mc + mt
Let's calculate acceleration which is common:
a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²
Now, when we know acceleration, let's calculate force on the truck:
Ft = mt · a = 2000 · 1.5 = 3000 N
Explanation:
Potential difference is measured in which units?
volts
amps
currents
watts
Answer:
Potential difference is measured in volts
Explanation:
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.
Answer:
Your answer is A.) volts
Explanation:
A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *
Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 2.60 m/s, grabs and holds onto him so that they move off together with a velocity of 1.30 m/s. If the mass of the tackler is 122 kg, determine the mass of the receiver. Assume momentum is conserved.
Answer:
122kgExplanation:
Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.
Momentum = Mass * Velocity
Before collision;
Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)
Momentum of the tackler
m2u2 = 2.60*122 = 317.2 kgm/s
where m2 and u2 are the mass and velocity of the tacker respectively.
Sum of momentum before collision = 0+317.2 = 317.2 kgm/s
After collision
Momentum of the bodies = (m1+m2)v
v = their common velocity
m1 = mass of the receiver
Momentum of the bodies = (122+m1)(1.30)
Momentum of the bodies = 158.6+1.30m1
According to the law above;
317.2 = 158.6+1.30m1
317.2-158.6 = 1.30m1
158.6 = 1.30m1
m1 = 158.6/1.30
m1 = 122kg
The mas of the receiver is 122kg
Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this
Answer:
Exothermic
Explanation:
Depending on the unit you are in, the answer may vary.
This is an exothermic reaction because it produces heat (the beaker gets warm).
When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Answer:
Fx = - (1/h)( wL2 + wL3 - wLend )
Explanation:
Assuming The twins Gilles and Jean has a weight ( w ) each
The torque that would balance the equation would be = wL2 + wL3 -------- 1
THEREFORE the ccw torques are = wLend + Fh ----------- 2
hence equation 2 equals equation 1
= wLend + Fh = wL2 + wL3 --------- 3
equation 3 can as well be represented as
F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4
From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative
therefore equation 4 is represented as
Fx = - (1/h)( wL2 + wL3 - wLend )
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference of 1.2 x 106 V and then enters a uniform magnetic field whose strength is 2.2 T. The alpha particle moves perpendicular to the field. Calculate (a) the speed of the alpha particle, (b) the magnitude of the magnetic force exerted on it, and (c) the radius of its circular path.
Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
[tex]K=qV[/tex] (1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):
[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]
The kinetic energy of the particle is also:
[tex]K=\frac{1}{2}mv^2[/tex] (2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]
the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:
[tex]|F_B|=qvBsin(\theta)[/tex]
B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1
[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]
the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:
[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]
the radius of the trajectory of the electron is 10.1 cm
The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m
What is the Magnetic force?
A) We know that the formula for kinetic energy can be expressed as;
K = qV
where;
q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C
V is potential difference = 1.2 × 10⁶ V
K = 3.2 × 10⁻¹⁹ * 1.2 × 10⁶
K = 3.84 × 10⁻¹³ J
Also, formula for kinetic energy is;
K = ¹/₂mv²
where v is speed
Thus;
v = √(2K/m)
v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)
v = 10.75 * 10⁶ m/s
B) The magnetic force is given by the formula;
F_b = qvB
F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)
F_b = 7.57 * 10⁻¹² N
C) The formula to find the radius is;
r = mv/qB
r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)
r = 0.101 m
Read more about magnetic field at; https://brainly.com/question/7802337
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly over its surface. find the magnitude and direction of the electric field (a) 2.2cm,(b)5.6cm,and (c)14 cm from the point charge.
Answer:
A) E = 278925.62 N/C with direction; radially out.
B) E = 43048.47 N/C with direction radially out.
C) E = -3214.29 N/C with direction radially in.
Explanation:
From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;
E = kQ/r²
where;
Q is the net charge within the distance r.
We are given the charge Q = 15-nC and
spherical shell of radius 10cm
A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
E = 278925.62 N/C
This will be radially out ,since the net charge is positive.
B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
E = 43048.47 N/C
This will be radially out ,since the net charge is positive.
C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Q = 15 nC - 22 nC
Q = -7 nC = -7 x 10^(-9) C
and;
E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
E = -3214.29 N/C
This will be radially in, since the net charge is negative. You can indicate this with a negative answer.
A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C
Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²Then E = 278925.62 N/CThen This will be radially out since the net charge is positive.
B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C
then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²Then E = 43048.47 N/CAfter that This will be radially out since the net charge is positive.
C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Then Q = 15 nC - 22 nCAfter that Q = -7 nC = -7 x 10^(-9) CWhen E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²Then E = -3214.29 N/C Thus, This will be radially in, since the net charge is negative.Find out more information about magnitude here:
https://brainly.com/question/13502329
A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
A worker pushes on a crate that experiences a net force of 45.0 N. If it accelerates at 0.500 m/s2 what is the weight?
Answer:
882 N
Explanation:
F = ma
45.0 N = m (0.500 m/s²)
m = 90.0 kg
mg = 882 N
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum heigh
Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
t = 3.74 s
(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
H = 136.86 m
(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
Vₓ = 41.83 m/s
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
Vy = 0 m/s
(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
ax = 0 m/s²
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
ay = 9.8 m/s²
An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?
Answer:
89.11kg
Explanation:
Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.
Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}
The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.
The weight of the fluid=0.087m3× 1kg/me = 0.087kg
Now the weight of the fluid displaced is referred to as the upthrust.
Now the real weight - the apparent weight = the upthrust.
Hence the apparent weight = real weight - upthrust
Apparent weight = 89.2-0.087 = 89.11kg
Jackson heads east at 25 km/h for 20 minutes before heading south at 45 km/h for 20 minutes. Hunter heads south at 45 km/h for 10 minutes before heading east at 40 km/h for 30 minutes. Find average velocity (magnitude and direction) of each person
Answer:
The average velocity of Jackson is 18.056 m/s South
The average velocity of Hunter is 10.65 m/s East
Explanation:
initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east
time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds
⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m
Final velocity of Jackson, v = 45 km/h South = 12.5 m/s South
time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes
time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m
Average velocity of Jackson;
[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]
initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South
time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds
⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m
Final velocity of Hunter, v = 40 km/h east = 11.11 m/s east
time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes
time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m
Average velocity of Hunter;
[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]
BEST ANSWER GETS BRAINLIEST!
At what distance from a 70.0 Watt speaker is the intensity 0.0195 W/m^2
(Treat the speaker as point of the source)
(Unit=meters)
PLEASE HELP ME!
Answer:
Distance = 16.9 m
Explanation:
We are given;
Power; P = 70 W
Intensity; I = 0.0195 W/m²
Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;
I = Power/Unit area = P/(4πr²)
where;
P is the sound power
r is the distance.
Thus;
Making r the subject, we have;
r² = P/4πI
r = √(P/4πI)
r = √(70/(4π*0.0195))
r = √285.6627
r = 16.9 m
Answer:
16.9 m
Explanation:
A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s
Answer:
B. 9.4 s
Explanation:
In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:
Vf₁ = Vi₁ + gt₁
where,
Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)
Vi = Initial Velocity in upward motion = 36.2 m/s
g = - 9.8 m/s² (negative due to upward motion)
t₁ = Time taken in upward motion = ?
Therefore,
0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)
t₁ = (36.2 m/s)/(9.8 m/s²)
t₁ = 3.7 s
Now, using 2nd equation of motion:
h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²
where,
h₁ = distance from top of building to highest point ball reaches = ?
Therefore,
h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²
h₁ = 133.58 - 66.86 m
h₁ = 66.72 m
No, considering downward motion and using subscript 2, for it.
Using 2nd equation of motion:
h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²
where,
h₂ = height of the highest point from ground = h₁ + height of building
h₂ = 66.72 m + 90 m = 156.72 m
Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)
t₂ = Time Taken in downward motion = ?
g = 9.8 m/s²
Therefore,
156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²
t₂² = (156.72 m)/(4.9 m/s²)
t₂ = √31.98 s²
t₂ = 5.7 s
Now, the total time taken by ball to reach the ground is"
Total Time = T = t₁ + t₂
T = 3.7 s + 5.7 s
T = 9.4 s
Therefore, the correct answer is:
B. 9.4 s
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?
Explanation:
The relation between resistance and resistivity is given by :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]\rho[/tex] is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
2. If rain is falling vertically downward, and you are running for shelter, should you hold your umbrella
vertically, tilted forward, or tilted backward to keep the driest? Please explain.
Answer:
Tilted forward to keep the driest.
Explanation:
The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.
The situation is the same as if you would stand still and the rain would come under an angle from the front.
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly windows conduct heat, calculate the rate of conduction in watts through a 2.82 m2 window that is 0.675 cm thick if the temperatures of the inner and outer surfaces are 5.00°C and −10.0°C, respectively. This rapid rate will not be maintained — the inner surface will cool, and frost may even form. The thermal conductivity of glass is 0.84 J/(s · m · °C).
Answer:
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat
Explanation:
Fourier's Law of heat conduction, gives the following formula:
Q = - KAΔT/t
where,
Q = Rate of Heat Conduction out of window = ?
K = Thermal Conductivity of Glass = 0.84 W/m.°C
A =Surface Area of window = 2.82 m²
ΔT = Difference in Temperature of both sides of surface
ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)
ΔT = 15°C
t = thickness of window = 0.675 cm = 0.00675 m
Therefore,
Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat.
A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 402 years
Explanation:
To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:
[tex]I=nqv_dA[/tex] (1)
n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3
q: charge of the electron = 1.6*10^-19 C
vd: drift velocity of electron in the metal = ?
A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2
I: current in the wire = 1110 A
You solve the equation (1) for vd:
[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]
Next, you calculate the time by using the information about the length of the line transmission:
[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]
hence, the electrons will take aproximately 402 years in crossing the line of transmission
John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the same kind of soup from 20 °C to 80 °C? The energy output per unit time by the heater is constant.
Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the ejection of photoelectrons. What changes will be observed if:
1. The silver is replaced with copper (Φ= 5.10 eV)?
a. more energetic photoelectrons (on average)
b. no photoelectrons are emitted more photoelectrons ejected
c. less energetic photoelectrons (on average)
d. fewer photoelectrons ejected
2. A second (identical) light source also shines on the metal?
a. fewer photoelectrons ejected
b. no photoelectrons are emitted more
c. energetic photoelectrons (on average)
d. less energetic photoelectrons (on average)
e. more photoelectrons ejected
3. The ultraviolet source is replaced with an X-ray source that emits the same number of photons per unit time as the original ultraviolet source?
a. no photoelectrons are emitted
b. less energetic photoelectrons (on average)
c. fewer photoelectrons ejected
d. more energetic photoelectrons (on average)
e. more photoelectrons ejected
Answer:
1. c
2. e
3. d
Explanation:
1.
From Einstein's Photoelectric Equation, we know that:
Energy given up by photon = Work Function + K.E of Electron
hc/λ = φ + K.E
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m
φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J
Therefore,
(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E
K.E = (9.939 - 8.16) x 10⁻¹⁹ J
K.E = 1.778 x 10⁻¹⁹ J
The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:
c. less energetic photo-electrons (on average)
2.
The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,
e. more photo-electrons ejected
3.
X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.
d. more energetic photo-electrons (on average)
Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules
Answer:
Hope this helps :)
Explanation:
1. A
2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)
For the parallel plates mentioned above, the DC power supply is set to 31.5 Volts and the plate on the right is at x = 14 cm. What is the magnitude of the electric field at a point on the x-axis where x = 7.0 cm? Answer with a number in the format ### in Newtons per Coulombs.
Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.
Answer:
E = 225 N/C
Explanation:
Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.
Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm
V(x) = 31.5 Volts
x = 14 cm = 0.14 m
The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:
E = V(x)/d
E(x = 0.14) = 31.5/0.14
E(x=0.14) = 225 N/C
E(x=0.14) = E(x=0.07) = 225 N/C
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is [tex]Q =2.094 C[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 0.205cm = 0.00205 \ m[/tex]
The radius of the wire is [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]
The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]
The electric field change is mathematically defied as
[tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]
Generally the charge is mathematically represented as
[tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]
So
[tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]
Therefore
[tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]
substituting values
[tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]
From the question we are told that t = 5 sec
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]
[tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]
[tex]Q =2.094 C[/tex]
The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.
Given the following data:
Diameter of wire = 0.205 centimeters.Resistivity of aluminum = [tex]2.75\times 10^{-8}[/tex] Ohm-meters.[tex]E(t)=0.0004t^2-0.0001t+0.0004[/tex] Newton per coulomb.Conversion:
Diameter of wire = 0.205 cm to m = 0.00205 meter.
Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]
To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:
First of all, we would find the area of the wire.
[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]
Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:
[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]
Where:
A is the area of a conductor.[tex]\rho[/tex] is the resistivity of a conductor.t is the time.E is the electric field.Substituting the given parameters into the formula, we have;
[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]
[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]
When t = 5 seconds:
[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]
Q = 2.094 Coulomb.
Find more information: https://brainly.com/question/18214726
In a 2 dimensional Cartesian system, the x-component of a vector is known, and the angle between vector and x-axis is known. Which operation is used to calculate the magnitude of the vector? (taken with respect to the x-component)
a. dividing by cosine
b. dividing by sine
c. multiplying by cosine
d. multiplying by sine
Answer:
The correct answer is a
Explanation:
The cosine function is
cos θ = ca / H
done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)
we clear
H = ca / cos θ
therefore, to find the magnitude of the vector, the cathete is divided into the cosine.
The correct answer is a