Answer:
C is the correct answer
Explanation:
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
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.
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
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:
A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed
Answer:
The new angular speed is [tex]w = 6 \ rev/s[/tex]
Explanation:
From the question we are told that
The angular velocity of the spin is [tex]w_o = 3 \ rev/s[/tex]
The original moment of inertia is [tex]I_o[/tex]
The new moment of inertia is [tex]I =\frac{I_o}{2}[/tex]
Generally angular momentum is mathematically represented as
[tex]L = I * w[/tex]
Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so
[tex]I * w = constant[/tex]
=> [tex]I_o * w _o = I * w[/tex]
where w is the new angular speed
So
[tex]I_o * 3 = \frac{I_o}{2} * w[/tex]
=> [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]
=> [tex]w = 6 \ rev/s[/tex]
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
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.
can I get help please?
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.
Answer:
Explanation:
Heat absorbed by oil
= mass x specific heat x rise in temperature
= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )
= 25.08 x 10⁵ J
Heat absorbed by air
= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )
= 6.03 x 10⁵ J
Total heat absorbed = 31.11 x 10⁵ J
If time required = t
heat lost from room
= .35 x 10³ t
Total heat generated in time t
= 1.8 x 10³ t
Heat generated = heat used
1.8 x 10³ t = .35 x 10³ t + 31.11 x 10⁵
1.45 x 10³ t = 31.11 x 10⁵
t = 31.11 x 10⁵ / 1.45 x 10³
t = 2145.5 s
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation:
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about point A, that is, the point of contact between the front tire of the forklift and the ground
Answer:
The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]
Explanation:
mass of the crate = [tex]M_{C}[/tex]
speed of forklift = [tex]V_{1}[/tex]
The distance between the center of the mass and the point A = d
Recall that the angular moment is the moment of the momentum.
[tex]L = P*d[/tex] ..... equ 1
where L is the angular momentum,
P is the momentum of the system,
d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image
Also, we know that the momentum P is the product of mass and velocity
P = mv ....equ 2
in this case, the mass = [tex]M_{C}[/tex]
the velocity = [tex]V_{1}[/tex]
therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]
we substitute equation 2 into equation 1 to give
[tex]L = M_{C} V_{1} d[/tex]
An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2. The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the direction of the cliff.
Required:
a. Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight.
b. The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the cliff the coyote lands.
c. Determine the components of the coyote’s impact velocity
Answer:
a) v_correcaminos = 22.95 m / s , b) x = 512.4 m ,
c) v = (45.83 i ^ -109.56 j ^) m / s
Explanation:
We can solve this exercise using the kinematics equations
a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff
v² = v₀² + 2 a x
they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²
v = √ (2 15 70)
v = 45.83 m / s
with this value calculate the time it takes to arrive
v = v₀ + a t
t = v / a
t = 45.83 / 15
t = 3.05 s
having the distance to the cliff and the time, we can find the constant speed of the roadrunner
v_ roadrunner = x / t
v_correcaminos = 70 / 3,05
v_correcaminos = 22.95 m / s
b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.
Let's start by looking for the time to reach the cliff floor
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
in this case y = 0 and the height of the cliff is y₀ = 100 m
0 = 100 + 45.83 t - ½ 9.8 t²
t² - 9,353 t - 20,408 = 0
we solve the quadratic equation
t = [9,353 ±√ (9,353² + 4 20,408)] / 2
t = [9,353 ± 13] / 2
t₁ = 11.18 s
t₂ = -1.8 s
Since time must be a positive quantity, the answer is t = 11.18 s
we calculate the horizontal distance traveled
x = v₀ₓ t
x = 45.83 11.18
x = 512.4 m
c) speed when it hits the ground
vₓ = v₀ₓ = 45.83 m / s
we look for vertical speed
v_{y} = [tex]v_{oy}[/tex] - gt
v_{y} = 0 - 9.8 11.18
v_{y} = - 109.56 m / s
v = (45.83 i ^ -109.56 j ^) m / s
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting
Answer:
Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m
Explanation:
The maximum electric field in the beam = 2E
cross-sectional area of beam = A
The intensity of an electromagnetic wave with electric field is
I = cε[tex]E_{0} ^{2}[/tex]/2
for [tex]E_{0}[/tex] = 2E
I = 2cε[tex]E^{2}[/tex] ....equ 1
where
I is the intensity
c is the speed of light
ε is the permeability of free space
[tex]E_{0}[/tex] is electric field
Radiation pressure of an electromagnetic wave on an absorbing surface is given as
P = I/c
substituting for I from above equ 1. we have
P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex] ....equ 2
Also, pressure P = F/A
therefore,
F = PA ....equ 3
where
F is the force
P is pressure
A is cross-sectional area
substitute equ 2 into equ 3, we have
F = 2ε[tex]E^{2}[/tex]A
force on a body = mass x acceleration.
that is
F = ma
therefore,
a = F/m
acceleration of the asteroid will then be
a = 2ε[tex]E^{2}[/tex]A/m
where m is the mass of the asteroid.
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct
Complete question:
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A. the interior field points in a direction parallel to the exterior field
B. There is no electric field on the interior of the conducting sphere.
C. The interior field points in a direction perpendicular to the exterior field.
D. the interior field points in a direction opposite to the exterior field.
Answer:
B. There is no electric field on the interior of the conducting sphere.
Explanation:
Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).
Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of free charges in the spherical conductor in response to any field until its neutralization.
Option B is correct.
There is no electric field on the interior of the conducting sphere.
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.
Question:
A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)
Answer:
3.22 x 10⁻⁴ m/s
Explanation:
The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;
v = [tex]\frac{I}{nqA}[/tex]
Where;
n = number of free electrons per cubic meter
q = electron charge
A = cross-sectional area of the wire
First let's calculate the number of free electrons per cubic meter (n)
Known constants:
density of copper, ρ = 8.95 x 10³kg/m³
molar mass of copper, M = 63.5 x 10⁻³kg/mol
Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol
But;
The number of copper atoms, N, per cubic meter is given by;
N = (Nₐ x ρ / M) -------------(ii)
Substitute the values of Nₐ, ρ and M into equation (ii) as follows;
N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³
N = 8.49 x 10²⁸ atom/m³
Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;
n = 8.49 x 10²⁸ electrons/m³
Now let's calculate the drift electron
Known values from question:
A = 5.261 mm² = 5.261 x 10⁻⁶m²
I = 23A
q = 1.6 x 10⁻¹⁹C
Substitute these values into equation (i) as follows;
v = [tex]\frac{I}{nqA}[/tex]
v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]
v = 3.22 x 10⁻⁴ m/s
Therefore, the drift electron is 3.22 x 10⁻⁴ m/s
Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Answer:
Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.
Explanation:
Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.
Answer and Explanation:
There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e. [tex]B \alpha \frac{1}{r^2}[/tex]
The magnetic field is a volume of vectors
And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit
Therefore for a closed path, we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
Observe the process by which the grey and the red spheres are charged using the electrophorus. After each sphere is first charged, what are their charges
Answer:
The gray spheres is negatively charged while the red is positively charged
Explanation:
This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged
Answer:
The gray sphere has a positive charge and the red sphere has a positive charge.
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12