Answer:
If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?
Charge will oscillate in the tank's capacitor and inductor.
Explanation:
A block is supported on a compressed spring, which projects the block straight up in the air at velocity . The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block? (Assume the ball can reach the block before the blochk reaches the ground and that the ball is thrown from a height equal to the release position of the block.)
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
B. At the instant when the block is at the highest point, directed at the block.
Explanation:
Motion of an object is the change in the position of the object with respect to time. On the earth, gravity has a great influence on the motion of an object (especially in a vertical direction).
When the block is projected up in the air, it moves with a varying velocity until the velocity becomes zero due to gravity. Which make the object to rest a little in the air (when velocity = gravity) and starts to fall freely.
To ensure hitting the block by the ball, it is thrown at the block when the block is at its highest point in the air. Since the block would be at rest at this instant before it start to fall at a constant acceleration under gravity.
Countries create quotas and tariffs to increase the volume of trade with their neighbors.
Oooooo, that statement is not true. Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.
Answer:
False
Explanation:
I took the text :)
How can socialism
impact populations?
Answer:
it represents a fundamental difference. (more info below)
Explanation:
Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.
hope this helped!
An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?
Answer:2000
Explanation:
Help with this answer please
Answer:
Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL
AAAAAAAAAAAA is the answer
HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?
Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.
What is the basis of soundproofing?Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.
Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.
Thus, the correct option is C, as the D image is showing the absorption.
For more details regarding soundproofing, visit:
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Answer: C.D
Explanation:...
when the same amount of heat is added to equal masses of water and copper at the same temperature the copper is heated to a higher final temperature than water. on a molecular level what explains this difference
a. the average kinetic energy of water molecules is greater than the average kinetic energy of the copper
b.more of the heat is transferred to the potential energy of the water molecules than the potential energy of the copper atoms
c.the intermolecular forces between copper atoms are stronger than those between water molecules
d.more of the heat is transferred to the kinetic energy of the water molecules than to the kinetic energy of the copper atoms
Answer:
C
Explanation:
The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.
Can someone help me with this question
Answer:
hypothesis , hope it helps
Explanation:
Answer:
Inference
Explanation:
Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.
Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.
Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
55.3
Explanation:
The computation of the number of bright-dark-bright fringe shifts observed is shown below:
[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]
where
d = [tex]3.95 \times 10^{-2}m[/tex]
[tex]\lambda = 400 \times 10^{-9}m[/tex]
n = 1.00028
Now placing these values to the above formula
So, the number of bright-dark-bright fringe shifts observed is
[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]
= 55.3
We simply applied the above formula so that the number of bright dark bright fringe shifts could come
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simultaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?
In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.
A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?
Answer:
A. F = 0.06 N
B. t = 6.37 s
Explanation:
A)
First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
where,
s = distance traveled = 1.62 m
Vi = 0 m/s (Since, it starts from rest)
t = Time Taken = 4.34 s
a = acceleration = ?
Therefore,
1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²
1.62 m/9.4178 s² = a
a = 0.172 m/s²
Now, from Newton's Second law, we know that:
F = ma
where,
F = Net Force of the combination = ?
m = Mass pf combination = 354 g = 0.354 kg
Therefore,
F = (0.354 kg)(0.172 m/s²)
F = 0.06 N
B)
Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:
F = ma
a = F/m
a = 0.06 N/0.762 kg
a = 0.08 m/s²
Now, using 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²
t² = 1.62 m/(0.04 m/s²)
t = √40.54 s²
t = 6.37 s
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?
Answer:
(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
[tex]I(t)=I_{max}sin(\omega t)[/tex] (1)
(a) The induced emf is given by the following formula
[tex]emf_L=-L\frac{dI}{dt}[/tex] (2)
You derivative the expression (1) in the expression (2):
[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
[tex]emf_L=-\omega LI_{max}[/tex]
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully
At t zero, the current through the inductor neither increasing nor decreasing because current is zero.
The current through an inductor of inductance L can be calculated by
[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1
(a) The induced emf can be calculated by
[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2
Derivative the equation (1) in the equation (2)
[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]
(b) At t=0 the current is zero,
(c) At t = 0 the emf is:
[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]
Therefore, at t zero, the current through the inductor neither increasing nor decreasing because current is zero.
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A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field B fills the region between the rails . Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :
A. v2m / ILB to yhe right
B. 3v2m /2 ILB to yhe left
C. 5v2m/ 2ILB to the right
D. v2m / 2ILB to the left
Answer:
F = ILB
Explanation:
To find the net force on the conducting bar you take into account the following expression:
[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]
I: current in the conducting bar
L: length of the bar
B: magnitude of the magnetic field
In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:
+i X +k = +j
The direction of the force is to the right and its magnitude is F = ILB
In each pair, select a substance that is a better heat conductor.
1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass
Answer:
1)copper wire
Explanation:
it is the best electric conductor
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.
Answer:
Letter A. [tex]y=4.55 m[/tex]
Explanation:
Let's use the wave equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction (x=1.21 m)Therefore the displacement y will be:
[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]
[tex]y=4.55 m[/tex]
The answer is letter A.
I hope it helps you!
Answer:
Explanation:
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
Amplitude (A) of the simple harmonic wave = 6.44 m
wave number (k) of the given wave = 2.34 m-1
Angular frequency (ω) of the given wave = 2.88 rad/s
Displacement x = 1.21 m and time t = 0.71 s
Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by
y = Asin(kx - ωt)
= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]
Y=6.44sin(0.7866 rad)
0.7866rad*(180 degrees/pi rad) =45.1
Y=6.44sin(45.1)
Y=4.55m
The amount of friction divided by the weight of an object forms a unit less number called the
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
[tex]F=\mu N[/tex]
N is normal force.
[tex]\mu[/tex] = coefficient of friction
[tex]\mu=\dfrac{F}{N}[/tex]
Question
20
what would be the advantages if your body had magnetic properties science subject
Answer:
Some of the advantages if our body had magnetic properties are as follows:
Magnetic properties can have health benefits such as recovering quickly from a stroke, resolving bladder problems, and reducing blood pressure.Brain will be able to control more activities of the nervous system and other organs of the body using magnetic power.Heart will have many benefits of magnetic properties and able to provide more energy to the entire body through the circulation of blood.Magnetic properties in body will be able to maintain the production of melatonin that controls the sleep patterns.Magnetic properties will be able to kill cancer causing cells.Hence, magnetic properties are somehow beneficial for humans.
A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?
Answer:
a. 63.89° in the north-southward manner
b. 2.2 sec
Explanation:
The goose is flying at 100 km/h
Air from east to west is 44 km/h
angle relative to the north-south direction for the bird to travel south will be
cos∅ = 44/100 = 0.44
∅ = [tex]cos^{-1}[/tex]0.44 = 63.89° in the north-southward manner
Speed south relative to the ground will be v
Tan 63.89 = v/100
2.04 = v/100
v = 2.04 x 100 = 204 km/hr
to cover a distance of 450 m from north to south at this speed time will be
t = d/v = 450/204 = 2.2 sec
A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.
What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.
Answer:
J = 0.800 kg m/s
Fmax = 291 N
Explanation:
During the fall, energy is conserved.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.81 m/s² × 1.45 m)
v = 5.33 m/s
Alternatively, you can use kinematics to find the velocity.
Impulse = change in momentum
J = Δp
J = mΔv
J = (0.150 kg) (5.33 m/s)
J = 0.800 kg m/s
Impulse = area under F vs t graph
J = ∫ F dt
J = ½ Fmax Δt
(0.800 kg m/s) = ½ Fmax (0.0055 ms)
Fmax = 291 N
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Answer:
The distance covered by puck A before collision is [tex]z = 8.56 \ m[/tex]
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is [tex]v_A = 3.90 \ m/s[/tex]
The speed of puck B is [tex]v_B = 4.30 \ m/s[/tex]
The distance covered by puck A is mathematically represented as
[tex]z = v_A * t[/tex]
=> [tex]t = \frac{z}{v_A}[/tex]
The distance covered by puck B is mathematically represented as
[tex]18 - z = v_B * t[/tex]
=> [tex]t = \frac{18 - z}{v_B}[/tex]
Since the time take before collision is the same
[tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]
substituting values
[tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]
=> [tex]70.2 - 3.90 z = 4.3 z[/tex]
=> [tex]z = 8.56 \ m[/tex]
A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.
Answer:
d. 1.00 m
Explanation:
In 1905, Einstein proposed special theory of relativity of light.
This theory had a number of consequences or results. One of them is called "Length Contraction".
According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.
But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.
In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:
d. 1.00 m
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.
Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s
Answer:
a) 98 m/s
b) 19.6 s
c) 1449.8 m
d) 29.6 s
e) 168.6 m/s
f) 46.8 s
Explanation:
Given that
Acceleration of the rocket, a = 5 m/s²
Altitude of the rocket, s = 960 m
a)
Using the equation of motion
v² = u² + 2as, considering that the initial velocity, u is 0. Then
v² = 2as
v = √2as
v = √(2 * 5 * 960)
v = √9600
v = 98 m/s
b)
Using the equation of motion
S = ut + ½at², considering that initial velocity, u = 0. So that
S = ½at²
t² = 2s/a
t² = (2 * 960) / 5
t² = 1920 / 5
t² = 384
t = √384 = 19.6 s
c)
Using the equation of motion
v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that
0 = 98² + 2(-9.8) * s
9600 = 19.6s
s = 9600/19.6
s = 489.8 m
The maximum altitude now is
960 m + 489.8 m = 1449.8 m
d)
Using the equation of motion
v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that
0 = 98 +(-9.8 * t)
98 = 9.8t
t = 98/9.8
t = 10 s
Total time then is, 10 + 19.6 = 29.6 s
e) using the equation of motion
v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,
v² = 0 + 2 * 9.8 * 1449.8
v² = 28416.08
v = √28416.08
v = 168.6 m/s
f) using the equation of motion
S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s
1449.8 = 0 + ½ * 9.8 * t²
2899.6 = 9.8t²
t² = 2899.6/9.8
t² = 295.88
t = √295.88
t = 17.2 s
total time in air then is, 17.2 + 29.6 = 46.8 s
When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir
(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Answer:
a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.
Explanation:
a) The change in entropy of the hot reservoir is:
[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]
[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]
b) The change in entropy of the cold reservoir is:
[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]
[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]
c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:
[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]
[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]
[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]
[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]
d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
b) The length of the rod:
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]
Given:
d = 1.5 mλ = 2.5 nC/m
Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.
Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]
How to calculate Electric Field?E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]
Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis
[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.
Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]
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brainly.com/question/14372859
You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 85 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
Force = 60.08 N
Explanation:
Given that
Diameter d = 30 mm
Holding pressure = 85 % of Atmospherics pressure
Solution
As we know that here 1 atm = 10⁵ N/m²
and pressure is known as force per unit area
pressure = [tex]\frac{F}{A}[/tex] ................1
put here value and we will get
F = [tex]0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N[/tex]
solve it we get
Force = 60.08 N
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
