Answer:
The rate of evaporation and the rate of condensation are the same.
Explanation:
I checked it myself on PennFoster
Hope This Helps!
A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that:
a. I1 > I2
b. I2 > I1.
c. I1 = I2.
Answer:
B: I2>I1
Explanation:
See attached file
A projectile is launched with V0 = 7.6 m/s and initial angle = 1.27 radians above the horizontal. What is the initial horizontal component of the projectile velocity in miles per hour?
Answer:
The horizontal component is [tex]v_h = 1.7096 \ m/s[/tex]
Explanation:
A diagram illustrating the projection is shown on the first uploaded image (from IB Maths Resources from British international school Phuket )
From the question we are told that
The initial velocity is [tex]v_o = 7.6 \ m/s[/tex]
The angle of projection is [tex]\theta = 1.27 \ rad = 72.77^o[/tex]
The horizontal component of this projectile velocity is mathematically represented as
[tex]v_h = v_o * cos (\theta )[/tex]
substituting values
[tex]v_h = 7.6 * cos (72.77 )[/tex]
[tex]v_h = 1.7096 \ m/s[/tex]
What is the increase in the internal energy of the system, if it absorbs 500 cal and at the same time does 85J work on its surroundings? (1 cal = 4.18J)
Answer:
2005 J
Explanation:
500 cal × (4.18 J/cal) = 2090 J
ΔU = q + w
ΔU = 2090 J − 85 J
ΔU = 2005 J
Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders
Answer:
From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
1. In a Millikan type experiment, two horizontal plates are 2.5 cm apart. A latex sphere of
mass 1.5 x 10-15 kg remains stationary when the potential difference between the
plates is 460 V, with the upper plate positive. [2+2+2+2 = 8 marks]
a. Is the sphere charged negatively or positively?
b. What is the magnitude of the electric field intensity between the plates?
C. Calculate the magnitude of the charge on the latex sphere.
d. How many excess or deficit electrons does the sphere have?
Answer:
Explanation:
a. Is the sphere charged negatively or positively?
The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.
Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.
b. What is the magnitude of the electric field intensity between the plates?
The magnitude of the electric field intensity between the plates is 18400v/m.
C. Calculate the magnitude of the charge on the latex sphere.
The magnitude of the charge on the latex sphere hae been solved and attached
d. How many excess or deficit electrons does the sphere have?
There are 5 excess electrons that the sphere has.
Check the attachment for further explanation.
Which best describes the relationship between heat, internal energy, and thermal energy?
Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
Mark this and return
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Submit
Answer:
I think it is the 4th answer choice
Explanation:
Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.
Inexperienced physics teachers often demonstrate the use of the electroscope by touching it with a charged glass rod at a single point. More experienced teachers typically drag the length of the rod across the top of the electroscope to increase the desired effect. Why does this help
Answer:
This is because The glass rod of the electroscope is an insulator therefore only charge transferred to the ball is at the point of contact on the rod. Thus, When the charge rod is dragged across the top of the electroscope, by the experienced teacher the more charge is transferred to electroscope thereby producing a greater effect
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Answer:
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Explanation:
Given;
orbital period of 3 years, P = 3 years
To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.
Kepler's third law;
P² = a³
where;
P is the orbital period
a is the orbital semi-major axis
(3)² = a³
9 = a³
a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]
Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
Answer:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
Explanation:
The magnitude of force applied by each charge on one another can be given by Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
where,
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now, in the final state the charges on both spheres are halved. Therefore,
q₁' = q₁/2
q₂' = q₂/2
Hence, the new force will be:
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
using equation 1:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
The magnitude of the electrostatic force will be F' = F/4
The magnitude of the electrostatic force:Here we used Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
Here
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now
q₁' = q₁/2
q₂' = q₂/2
So, the new force should be
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
So,
F' = F/4
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A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes. How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
A. 0.438c
B. 0.671c
C. 0.319c
D. 0.874c
E. 0.658c
The spaceship is traveling towards the Sun with velocity 0.319c.
What is special theory of relativity?
One of the most significant works in the history of physics is Albert Einstein's 1905 theory of special relativity. The theory of special relativity explains how speed affects space, time, and mass. Small amounts of mass (m) can be interchangeable with large amounts of energy (E), as defined by the classic equation E = mc², according to the theory, which offers a means for the speed of light to define the link between energy and matter.
Given parameters:
The radius of the orbit of Jupiter is 43.2 light-minutes.
that of the orbit of Mars is 12.6 light-minutes.
Hence, actual distance between them l₀= 43.2 - 12.6 light-minute = 30.6 light-minute.
Let the velocity of the spaceship is v.
Time taken to travel this actual distance = l₀ √(1+v²/c²)/v
According to the question
l₀ √(1+v²/c²)/v = 35.0
√(c²+v²) = 35.0 v
c² = 1224 v²
v = 0.319c
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A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Answer:
3.09 m/s²
Explanation:
Given:
Δx = -105 m
v₀ = -27.7 m/s
v = -10.9 m/s
Find: a
v² = v₀² + 2aΔx
(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)
a = 3.09 m/s²
A parallel-plate capacitor is formed from two 9.1 cm-diameter electrodes spaced 1.3 mm apart. The electric field strength inside the capacitor is 5.0×106N/C.
What is the charge (in nC) on each electrode?
Answer:
2.87nC
Explanation:
See attached file
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as
Question:
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as θ = 2.70t² − 0.900t3³
where θ is in radians and t is in seconds.
(a) Find the maximum angular speed of the roller.
(b) What is the maximum tangential speed of a point on the rim of the roller?
(c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation?
(d) Through how many rotations has the roller turned between t = 0 and the time found in part (c)?
Answer:
a. 2.7rad/s
b. 8.1 m/s
c. 2 s
d. 3.6 rad
Explanation:
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters. What is the magnitude of the tension on the ends of the clothesline
Answer:
The tension on the clotheslines is [tex]T = 8.83 \ N[/tex]
Explanation:
The diagram illustrating this question is shown on the first uploaded image
From the question we are told that
The distance between the two poles is [tex]d = 12 \ m[/tex]
The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]
The length at which the clotheslines sags is [tex]l = 4 \ m[/tex]
Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as
[tex]W = mg[/tex]
let the angle which the tension on the clotheslines makes with the horizontal be [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows
[tex]Tan \theta = \frac{ 4}{6}[/tex]
=> [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]
=> [tex]\theta = 33.70^o[/tex]
So the vertical component of this tension is mathematically represented a
[tex]T_y = 2* Tsin \theta[/tex]
Now at equilibrium the net horizontal force is zero which implies that
[tex]T_y - mg = 0[/tex]
=> [tex]T sin \theta - mg = 0[/tex]
substituting values
[tex]T = \frac{m*g}{sin (\theta )}[/tex]
substituting values
[tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]
[tex]T = 8.83 \ N[/tex]
A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 {\rm mi} away. He travels at a steady 47.0 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 55.0 mph. Who gets to San Francisco first? How long does the first to arrive have to wait for the second?
Complete question is;
Alan leaves Los Angeles at 8:AM to drive to San Francisco, 400 mi away. He travels at a steady 47 mph. Beth leaves Los Angeles at 9:00 AM and drives a steady 55.0 mph.
a. Who gets to San Francisco first?
b. How long does the first to arrive have to wait for the second?
Answer:
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
Explanation:
We are given;
Distance between Los Angeles and San Francisco;d = 400 mi
Alan's velocity;v_a = 47 mph
Beth's velocity;v_b = 55 mph
we know that time is given by;
Time = distance/velocity
Time required for alan;
t_a = 400/47
t_a = 8.5106 hours = 8 hours 31 minutes
Time required for Beth;
t_b = 400/55
t_b = 7.2727 hours = 7 hours 16 minutes
So Alan will reach there at;
8:00 a.m + 8 hours 31 minutes = 4:31 p.m
Beth will reach there at;
9:00 a.m + 7 hours 16 minutes = 4:16 pm
Thus, beth will arrive 4:31 pm - 4:16pm = 15 minutes before alan
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
A diver running at 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 seconds later reaches the water below. How far from its base did the diver hit the water
Explanation:
u = 2.5 m/s
v = 0
t = 3sec
s = ?
s = (u+v)/t
s = (0+2.5)/3
s = 2.5/3 = 0.83 m
Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the outer sphere is 1.62 10-8 C. Find the magnitude of the electric field (in N/C) at 0.012 m.
Answer:
The electric field is [tex]E = 2.2625 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner sphere is [tex]r_1 = 0.008\ m[/tex]
The radius of the outer sphere is [tex]r _2 = 0.018 \ m[/tex]
The charge on the inner sphere is [tex]q_1 = 3.62 *10^{-8} \ C[/tex]
The charge on the outer sphere is [tex]q_2 = 1.62 *10^{-8} \ C[/tex]
The position from the origin is [tex]d = 0.012 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k (q_1 )}{ r^2}[/tex]
The reason for using [tex]q_1[/tex] for the calculation is due to the fact that the position considered is greater than the [tex]r_1[/tex] but less than [tex]r_2[/tex]
Here k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}[/tex]
So
[tex]E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}[/tex]
[tex]E = 2.2625 *10^{6} \ N/C[/tex]
A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the positive direction.
(a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car .
(b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case 33%
(c) What is the ratio of the force on the car from the concrete to the braking force?
Answer:
A) Force = 2303.925 N in the negative x-direction
B) F ≈ 143998.28 N
C) Ratio = 62.5
Explanation:
A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.
Let's find the deceleration using Newton's law of motion formula;
v² = u² + 2as
where;
v = final velocity,
u = initial velocity,
s = displacement
a = acceleration
We are given;
u = 87.5 km/h = 24.3056 m/s
s = 125 m
v = 0 m/s
Thus;
0 = (24.3056)² + 2a(125)
- (24.3056)²= 250a
a = - 24.3056²/250
a = - 2.363 m/s²
Now, force = mass × acceleration
We are given mass = 975 kg
Thus;
Force = 975 x (-2.363)
Force = 2303.925 N in the negative x-direction
B) formula for kinetic energy is
KE = ½mv²
KE = ½(975)(24.3056)²
= 287996.568288 J
Work done on impact = F x 2
Thus;
2F = 287996.568288
F = 287996.568288/2
F ≈ 143998.28 N
C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated volume? Which method might be more accurate? Why?
Answer:
The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate
Explanation:
This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
Suppose you take two non-zero displacements represented by vectors A & B.The magnitude of A is 5 m and the magnitude of B is 6 m. Under what circumstances can you end up back at your starting point? What is the magnitude of the largest displacement you can end up from the starting point?
Suppose you take two non-zero displacements represented by vectors A & B which are perpendicular to each other. The magnitude of A is 5 m and the magnitude of B is 6 m. What is the component of vector B along the direction of vector A measured in m?
Answer:
a. When the total displacement is -(A + B)
b. A + B = 1 m or -(A + B) = -11 m
c. 0 m
Explanation:
a. Under what circumstances can you end up back at your starting point?
If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B
b. What is the magnitude of the largest displacement you can end up from the starting point?
The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.
c. When A and B are perpendicular, what is the component of B in the direction of A?
Since A is perpendicular to B, the angle between A and B is 90°
So the component of B in A,s direction is Bcos90° = B × 0 = 0 m
difference between freefall and weightlessness.
Answer:
Differences between freefall and weightlessness are as follows:
FreefallWhen a body falls only under the influence of gravity, it is called free fall.Freefall is not possible in absence of gravity.A body falling in a vacuum is an example of free fall.WeightlessnessWeightlessness is a condition at which the apparent weight of body becomes zero.Weightlessness is possible in absence of gravity.A man in a free falling lift is an example of weightlessness.Hope this helps....
Good luck on your assignment....
How does an atom of rubidium-85 become a rubidium ion with a +1 charge?
Answer:
C. The atom loses 1 electron to have a total of 36.
Explanation:
Cations have a positive charge. Cations lose electrons.
The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.
You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is gold (rhog=2.44×10^−8Ω⋅m), one is copper (rhoc=1.72×10^−8Ω⋅m), and one is aluminum (rhoa=2.75×10−8Ω⋅m).
Required:
a. What will be the length of the gold wire?
b. What will be the length of the copper wire?
c. What will be the length of the aluminum wire?
d. Gold has a density of 1.93 × 10^4 kg/m^3. What will be the mass of the gold wire?
e. If gold is currently worth $40 per gram, what is the cost of the gold wire?
Answer:
(a) L = 128.75 m
(b) L = 182.56 m
(c) L = 114.28 m
(d) Mass of Gold = 7.68 kg = 7680 gram
(e) Cost of Gold Wire = $ 307040
Explanation:
The resistance of the wire is given as:
R = ρL/A
where,
R = Resistance
ρ = resistivity
L = Length
A = cross-sectional area
(a)
For Gold Wire:
ρ = 2.44 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)
L = 128.75 m
(b)
For Copper Wire:
ρ = 1.72 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)
L = 182.56 m
(c)
For Aluminum Wire:
ρ = 2.75 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)
L = 114.28 m
(d)
Density = Mass/Volume
Mass = (Density)(Volume)
Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³
Therefore,
Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)
Mass of Gold = 7.68 kg = 7680 gram
(e)
Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)
Cost of Gold Wire = ($ 40/gram)(7680 grams)
Cost of Gold Wire = $ 307040
(a) L is = 128.75 m
(b) L is = 182.56 m
(c) L is = 114.28 m
(d) Mass of Gold is = 7.68 kg = 7680 gram
(e) Cost of Gold Wire is = $307040
Calculation of Diameter cylindricalWhen The resistance of the wire is given as:
R is = ρL/A
Now, where
R is = Resistance
ρ is = resistivity
L is = Length
A is = cross-sectional area
(a) For Gold Wire is:
ρ is = 2.44 x 10⁻⁸ Ω.m
A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
1 Ω is = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L is = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)
L is = 128.75 m
(b) For Copper Wire is:
ρ is = 1.72 x 10⁻⁸ Ω.m
Then A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
After that 1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
Now, L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)
Therefore, L = 182.56 m
(c) For Aluminum Wire is:
ρ is = 2.75 x 10⁻⁸ Ω.m
A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
Then 1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
After that L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)
L = 114.28 m
(d) Density is = Mass/Volume
Mass is = (Density)(Volume)
Then Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³
Therefore,
Now, Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)
Then Mass of Gold = 7.68 kg = 7680 gram
(e) The Cost of Gold Wire is = (Unit Price of Gold)(Mass of Gold)
Than Cost of Gold Wire = ($ 40/gram)(7680 grams)
Therefore, The Cost of Gold Wire is = $ 307040
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A particle on the x-axis is moving to the right at 2 units per second. At a certain instant it is at the point (5,0). How rapidly it the distance between the particle and the point (0,9) on the y-axis changing at that instant
Answer:
dr/dt = 1.94 units per second
Explanation:
A particle is moving on the x-axis to the right at 2 u/s.
To know how is changing the distance of the particle respect to the point (0,9), on the y-axis, you first take into account that the distance between charge and a point over the y-axis is given by:
[tex]r^2=x^2+y^2[/tex] (1)
Next, you calculate implicitly the derivative of the equation (1) respect to t:
[tex]\frac{d}{dt}r^2=\frac{d}{dt}[x^2+y^2]\\\\2r\frac{dr}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}[/tex] (2)
Next, you solve the previous equation for dr/dx:
[tex]\frac{dr}{dt}=\frac{x(dx/dt)+y(dy/dt)}{\sqrt{x^2+y^2}}[/tex] (3)
dx/dt: the speed of the particle on the x-axis = 2
dy/dt: speed of the particle on the y-axis = 0
For the instant given in the statement, you have that:
x = 5
y = 9
Then, you replace the values for x, y, dx/dt and dy/dt in the equation (3):
[tex]\frac{dr}{dt}=\frac{2(5)(2)+0}{\sqrt{(5)^2+(9)^2}}=1.94[/tex]
The speed of change of the distance between particle and point (0,9) is 1.94 units per second
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]