Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
Binding Energy = 2.24 eV
Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet
Answer:
A. magnetic field
Explanation:
The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .
What is a magnetic field ?A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.
As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.
The magnetic field is produced as a result when an electrical current is passed through the conducting wire .
Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .
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Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?
Answer: Only Tech B is correct.
Explanation:
First, tech A is wrong.
The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.
Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.
You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb?
a. at the point, because all rays bouncing off the mirror will be parallel.
b. at the focal point of the mirror
c. at the radius of curvature of the mirror
d. none of the above, you cant make parallel rays wilth a concave mirror
Answer:
Explanation:
Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.
For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.
Therefore, the light bulb should be placed at the focal point of the mirror.
In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?
Answer:4 times more energy will be striking the childbearing
Explanation:
Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4
A ferry boat sails east across a lake at 10 km/h. A woman is walking east on
the boat at 1.5 km/h. What is her speed relative to the boat?
A. 8.5 km/h west
B. 8.5 km/h east
C. 1.5 km/h east
O D. 1.5 km/h west
Answer:
B
8.5 km/h east
Explanation:
Relative velocity= Va -Vb
=10-1.5
=8.5 km/h east
The concept relative speed is used when two or more bodies moving with some speed are considered. The relative speed of woman to the boat is 8.5 km/h east. The correct option is B.
What is relative speed?The relative speed of two bodies is defined as the sum of their speeds if they are moving in the opposite direction and it is the difference of their speeds if they are moving in the same direction.
The speed of the moving body with respect to the stationary body is known as the relative speed. The term relative means in comparison to. The relative speed is a scalar quantity.
Here both the boat and women are travelling in the same direction. So the relative speed is given as:
Relative speed = 10 - 1.5 = 8.5 km / h
Therefore the relative speed is 8.5 km/h east.
Thus the correct option is B.
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Unpolarized light is passed through three successive Polaroid filters, each with its transmission axis at 45.0° to the preceding filter. What percentage of light gets through?
Answer:
The percentage is [tex]k = 12.5 \%[/tex]
Explanation:
From the question we are told that
The axis is is at [tex]\theta = 45 ^o[/tex]
Generally the of intensity light emerging from the first polarizer is mathematically represented as
[tex]I_{1} = \frac{I_o}{ 2}[/tex]
Where [tex]I_o[/tex] is the intensity of unpolarized light
Now the light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]
[tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]
[tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]
Now the light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]
[tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Now the percentage of the intensity of light that emerged with respect to the intensity of the unpolarized light is
[tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]
[tex]k = 12.5 \%[/tex]
The percentage of light that gets through the three successive Polaroid filters is; 12.5%
We are given;
Angle of transmission axis; θ = 45°
Formula for intensity of light from first polarizer is;
I₁ = ¹/₂I₀
Formula for intensity of light from second polarizer is;
I₂ = I₁cos²θ
Formula for intensity of light from third polarizer is;
I₃ = I₂cos²(90 - θ)
Combining the 3 equations;
Put ¹/₂I₀ for I₁ in second formula to get;
I₂ = ¹/₂I₀cos²θ
Put ¹/₂I₀cos²θ for I₂ in third formula to get;
I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)
Plugging in 45° for θ gives;
I₃ = ¹/₂I₀cos²45*cos²(90 - 45)
⇒ I₃ = ¹/₂I₀cos²45*cos²45
⇒ I₃ = ¹/₂I₀cos⁴45
Now, cos 45 in surd form is 1/√2. Thus;
I₃ = ¹/₂I₀(1/√2)⁴
I₃ = ¹/₂I₀(¹/₄)
I₃ = ¹/₈I₀
I₃/I₀ = ¹/₈
I₃/I₀ = 0.125
In percentage form, we have;
I₃/I₀ = 12.5%
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A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg . Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.
Answer:
602.27 kg
Explanation:
The computation of the largest mass of cargo the balloon can lift is shown below:-
Volume of helium inside the ballon= (4 ÷ 3) × π × r^3
= (4 ÷ 3) × 3.14 × 6.953
= 1406.19 m3
Mass the balloon can carry = volume × (density of air-density of helium)
= 1406.19 × (1.29-0.179)
= 1562.27 kg
Mass of cargo it can carry = Mass it can carry - Mass of structure
= 1562.27 - 960
= 602.27 kg
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastward. What are the magnitude (in T) and direction of the magnetic field at this instant?
Answer:
The values is [tex]B = 3.2 *10^{-8} \ T[/tex]
The direction is out of the plane
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 9.6 \ V/m[/tex]
The magnitude of the magnetic field is mathematically represented as
[tex]B = \frac{E}{c}[/tex]
where c is the speed of light with value
[tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]
[tex]B = 3.2 *10^{-8} \ T[/tex]
Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
Answer:
a
[tex]m = 0.169 \ kg[/tex]
b
[tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 14 \ N/m[/tex]
The maximum extension of the spring is [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
The number of oscillation is [tex]n = 30[/tex]
The time taken is [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
[tex]T = \frac{t}{n}[/tex]
substituting values
[tex]T = \frac{20}{30 }[/tex]
[tex]T = 0.667 \ s[/tex]
Thus
[tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
[tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]m =\frac{k }{w^2}[/tex]
substituting values
[tex]m =\frac{ 15 }{(9.421)^2}[/tex]
[tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is mathematically represented as
[tex]v = - Awsin (wt)[/tex]
The velocity is maximum when [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
[tex]v_{max} = - A* w[/tex]
=> [tex]|v_{max} |= A* w[/tex]
=> [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> [tex]|v_{max} |= 0.5653 \ m/s[/tex]
The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)
Answer:
Explanation:
Let f₀ be the frequency .
energy of photons having frequency of f₀
= hf₀ where h is plank's constant
for electron to get ejected , work function should be equal to energy of photon
hf₀ = 3.55
4.1357 x 10⁻¹⁵ x f₀ = 3.55
f₀ = 8.58 x 10¹⁴ Hz .
A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the magnetic flux through the loop
Answer:
The magnitude of the magnetic flux through the loop is 0.0982 T.m²
Explanation:
Given;
magnitude of magnetic field, B = 0.5 T
radius of the loop, r = 0.25 m
Area of the loop is given by;
A = πr²
A = 3.142 x (0.25)²
A = 0.1964 m²
The magnitude of the magnetic flux through the loop is given by;
Ф = BA
Where;
B is the magnitude of the magnetic field
A is area of the field
Ф = 0.5 x 0.1964
Ф = 0.0982 T.m²
Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?
Answer:
1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216
Explanation:
In the double-slit interference phenomenon it is explained for constructive interference by the equation
d sin θ = m λ
1) the first order maximum occurs for m = 1
sin θ = λ / d
θ = sin⁻¹ λ / d
let's reduce the magnitudes to the SI system
λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm
d = 0.440 mm = 0.440 10⁻³ m ³
let's calculate
θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)
θ = sin⁻¹ (1.18 10⁻³)
θ = 0.00118 rad
2) the second order maximum occurs for m = 2
θ = sin⁻¹ (m λ / d)
θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)
θ = 0.00236 rad
3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains
I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )
where the function sinc = sin x / x
and b is the width of the slits
we caption the values
x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)
x = 2.21
I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²
remember angles are in radians
I / I₀ = cos² (3.0945) [0.363] 2
I / I₀ = 0.9978 0.1318
I / I₀ = 0.1738
4) the maximum second intensity is
I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)
x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)
x = 4.41
I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²
I / Io = cos² 6.273 0.216
I / Io = 0.216
.
Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.
Answer:
A) 1.5 V
B) 4.5 V
Explanation:
A) Batteries in parallel have the same voltage as an individual battery.
V = 1.5 V
B) Batteries in series have a voltage equal to the sum of the individual batteries.
V = 1.5 V + 1.5 V + 1.5 V
V = 4.5 V
An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?
Answer:
3.9E-8
Explanation:
We know that
Mv²/r = Bqv
So
r= mv/Bq
But E is 1/2mv² which is 5.53eV
m²v² =2m x 5.53eV
mv = √( 2 x 9.1E-31)
So
r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19
= 3.9x10-8cm
Explanation:
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits. Find the distance along the screen between adjacent bright fringes.
Answer:
The distance is [tex]y = 0.00162 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]
The slit spacing is [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 2.0 \ m[/tex]
Generally the distance along the screen between adjacent bright fringes is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]
[tex]y = 0.00162 \ m[/tex]
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the other pipe?
Answer:
The length of the longer pipe is L = 2.30 m
Explanation:
Given that:
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.
How long is the other pipe?
From above;
The formula for the frequency of open ended pipes can be expressed as:
[tex]f = \dfrac{nv}{2L}[/tex]
where n = 1 ( since half wavelength exist between those two pipes)
v = 343 m/s and L = 2.08 m
Thus, the shorter pipe produces a frequency of :
[tex]f = \dfrac{1*343}{2*2.08}[/tex]
[tex]f = \dfrac{343}{4.16}[/tex]
[tex]f =82.45 \ Hz[/tex]
Also; we know that the beat frequency was given as 8.0 Hz
Then,
The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz
The lower frequency of the longer pipe = 74.45 Hz
Finally;
From the above equation; make Length L the subject of the formula. Then,
The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]
The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]
The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]
The length of the longer pipe is L = 2.30 m
What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.
Answer:
C
They allow light to pass only if their directions of polarizations are exactly 90° apart.
A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizontal. If the initial speed of the stream is 40 m/s the height that the stream of water will strike the building is
Answer:
We can think the water stream as a solid object that is fired.
The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)
The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).
The initial speed of the stream is 40m/s.
First, using the fact that:
x = R*cos(θ)
y = R*sin(θ)
in this case R = 40m/s and θ = 30°
We can use the above relation to find the components of the velocity:
Vx = 40m/s*cos(30°) = 34.64m/s
Vy = 20m/s.
First step:
We want to find the time needed to the stream to hit the buildin.
The horizontal speed is 34.64m/s and the distance to the wall is 50m
So we want that:
34.64m/s*t = 50m
t = 50m/(34.64m/s) = 1.44 seconds.
Now we need to calculate the height of the stream at t = 1.44s
Second step:
The only force acting on the water is the gravitational one, so the acceleration of the stream is:
a(t) = -g.
g = -9.8m/s^2
For the speed, we integrate over time and we get:
v(t) = -g*t + v0
where v0 is the initial speed: v0 = 20m/s.
The velocity equation is:
v(t) = -g*t + 20m/s.
For the position, we integrate again over time:
p(t) = -(1/2)*g*t^2 + 20m/s*t + p0
p0 is the initial height of the stream, this data is not known.
Now, the height at the time t = 1.44s is
p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po
= 16.57m + p0
So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?
Answer:
c. 70 Ω
Explanation:
The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:
Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)
Req = R/2 + 2R/2 + 4R/2
Req = 3.5R
Req = 70Ω
From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:
a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down
Answer:
c. 1.11 m/s down
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Assuming the balloon and projectile are originally at rest:
(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)
0 kg m/s = (90 kg) v + 100 kg m/s
v = -1.11 m/s
An inductor is hooked up to an AC voltage source. The voltage source has EMF V0 and frequency f. The current amplitude in the inductor is I0.
Part A
What is the reactance XL of the inductor?
Express your answer in terms of V0 and I0.
Part B
What is the inductance L of the inductor?
Express your answer in terms of V0, f, and I0.
Answer:
a. The reactance of the inductor is XL = V₀/I₀
b. The inductance of the inductor is L = V₀/2πfI₀
Explanation:
PART A
Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,
XL = V₀/I₀
So, the reactance of the inductor is XL = V₀/I₀
PART B
The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor
Since XL = V₀/I₀ = 2πfL
V₀/I₀ = 2πfL
L = V₀/2πfI₀
So the inductance of the inductor is L = V₀/2πfI₀
A) The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex]
B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
A) Expressing the Reactance of the inductor
Voltage across the Inductor = V₀ = I₀XL ---- ( 1 )
Where : V₀ = emf voltage , I₀ = current
from equation ( 1 )
∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]
B ) Expressing the Inductance of the Inductor
Inductance of an inductor is expressed as : XL = 2πfL
from part A
XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL
∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀
L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
Hence we can conclude that The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex] and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex] .
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The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above
Answer:
b. Third Law entropy
Explanation:
Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature" tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.
In the question above, the correct answer is option b.
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?
Answer:
The velocity is [tex]v_h = 19.2 \ m/s[/tex]
Explanation:
From the question we are told that
The speed of the roller coaster at ground level is [tex]v = 26 \ m/s[/tex]
Generally we can define the roller coaster speed at ground level using the an equation of motion as
[tex]v^2 = u^2 + 2 g s[/tex]
u is zero given that the roller coaster started from rest
So
[tex]26^2 = 0 + 2 * g * s[/tex]
So
[tex]s = \frac{26^2}{ 2 * g }[/tex]
=> [tex]s = 37.6 \ m[/tex]
Now the displacement half way is mathematically represented as
[tex]s_{h} = \frac{37.6}{2}[/tex]
[tex]s_{h} = 18.8 \ m[/tex]
So
[tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]
Where [tex]v_h[/tex] is the velocity at the half way point
=> [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]
=> [tex]v_h = 19.2 \ m/s[/tex]
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
[tex]Em_{f}[/tex] = U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h
Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.
Answer:
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Explanation:
In this exercise you are asked to relate each with the answers
In general, in the optics diagram,
* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point
* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate
* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.
With these statements, let's review the answers
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Find an analytic expression for p(V)p(V)p(V), the pressure as a function of volume, during the adiabatic expansion.
Answer:
In an adiabatic process we have
pV γ = const..
This explains that the pressure is a function of volume, p ( V ) ,
So can be written as:
p ( V ) × V γ = p 0 V γ 0 ,
or p ( V ) = p 0 V 0 / V γ
= p 0 V 0 / V ^(7 / 5)
