The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
[tex]A = 10 \ cm^2[/tex]
[tex]$=0.0010 \ m^2$[/tex]
d = 10 mm
= 0.010 m
Then, Capacitance,
[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]
[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]
[tex]$C=2.655 \times 10^{12} \ F$[/tex]
[tex]$U_1 = \frac{1}{2}CV^2$[/tex]
[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]
Now,
[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]
And
[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]
In parallel combination,
[tex]$C_{eq}= C_k + C_{air}$[/tex]
[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]
[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]
[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]
Then energy,
[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]
[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]
b). Now the charge on the [tex]\text{capacitor}[/tex] is :
[tex]$Q=C_{eq} V$[/tex]
[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]
[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]
Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :
[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]
[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]
[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]
[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]
Without the dielectric,
[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]
[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]
[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]
Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?
A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J
=F×s×cosa=2×g×0,8×cos90°= 0
The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.
What is Work?Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.
Work = Force × Displacement
Force = Mass × Acceleration
Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.
Therefore, the correct option is A.
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Explain how newton's first law of motion follows from second law?
Answer:
Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.
Which of the following is a form of mechanical energy?
A. Chemical energy
B. Gravitational potential energy
C. Thermal energy
D. Nuclear energy
Answer:
B
Explanation:
no reason for this answer
5. A bicyclist is finishing her repair of a flat tire when a friend rides by at a constant velocity of
3.5 m/s. Three seconds later, the bicyclist hops on her bike and accelerates at 3.6 m/s² until she
catches her friend.
a. How much time does it take until she catches her friend?
b. How far has she traveled in this time?
c. What is her speed when she catches up?
Answer:
a) t = 3.6 s
b) d = 23 m
c) v = 13 m/s
Explanation:
Let t be the time the accelerating rider rides
the distance she travels is
d = ½3.6t²
the distance for the other cyclist is
d =3.5(t + 3)
½3.6t² = 3.5(t + 3)
1.8t² - 3.5t - 10.5 = 0
quadratic formula, positive answer
t = (3.5 + √(3.5² - 4(1.8)(-10.5))) / (2(1.8))
t = 3.575786...
d = ½(3.6)(3.575786²) = 23.015...
v = 3.6(3.575786) = 12.8728...
1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.
Answer:
B
Explanation:
P1/P1 = 40/20
=2
A visible violet light emits light with a wavelength of 4.00 × 10-7 m.
Calculate the frequency of the violet light.
A)6.30 × 10 -1 Hz
B)7.50 × 10 14 Hz
C)6.30 × 10 24 Hz
D)7.50 × 10 1 Hz
Answer:
The correct option is B. 7.5 * 10¹⁴ Hz
Explanation:
Frequency = (speed) / (wavelength)
= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)
= (3/4 x 10¹⁵) ( m / m - s )
= (0.75 x 10¹⁵) /sec
= 7.5 x 10¹⁴ Hz
= 750,000 GHz
Answer:
Mark Brainliest please
answer is
Explanation:
For any wave,
Frequency = (speed) / (wavelength)
= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)
= (3/4 x 10¹⁵) ( m / m - s )
= (0.75 x 10¹⁵) /sec
= 7.5 x 10¹⁴ Hz
= 750,000 GH
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
Dựa vào môi trường hoạt chất, laser được phân thành
Answer:
ok
Explanation:
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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A person runs up the stairs elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.
Determine the work done by the person climbing the stair case.
Answer:
Work done = 2289.084 Joules
Explanation:
Given the following data;
Mass = 102 Kg
Height = 2.29
Time = 1.32 seconds
We know that acceleration due to gravity, g = 9.8 m/s²
a. To find the work done by the person;
Here, work would be done in the form of gravitational potential energy.
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;
G.P.E = mgh
Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the formula, we have;
Work done = 102 * 2.29 * 9.8
Work done = 2289.084 Joules
What is the percentage of the population that wanted both the swimming pool and the soccer complex? Use your knowledge
of the addition rule and the Venn diagram to answer.
Answer:
The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.
Explanation:
Edmentum
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
A smokestack of height H = 50 m emits a pollutant in a 3 m/s wind. The plume is carried downwind by advection (wind speed U = 3 m/s) and is simultaneously dispersing vertically with a turbulent diffusion coefficient D. The vertical diffusion causes the plume to widen vertically over time, with halfâwidth (distance from centerline to edge) increasing as:
half width = 2 â2Dt
The plume reaches the ground some distance L downwind of the base of the smokestack (see sketch in book on page 203)
a. If L = 2 km, estimate the value of the turbulent diffusion coefficient D.
b. Under the same wind speed and turbulence conditions, what would be the value of L if the smokestack were twice as high?
Answer:
a) 0.46875
b) 8 km
Explanation:
Smokestack height ( H ) = 50 m
speed of pollutant / wind speed = 3 m/s
Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m ---- ( 1 )
a) If L = 2 km
value of turbulent diffusion coefficient D
back to equation 1
50 = 2 √ 2 * D * ( 2000/3 )
2500 = 4 * 2 * D * ( 2000/3 )
D = 2500 / ( 8 * ( 2000/3 ) )
= 0.46875
where : time to travel ( t ) = Distance / speed = 2000 / 3
b) when the smoke stack = 50 * 2 = 100 m
L = 800 m = 8 km
attached below is the detailed solution
Partial tides _______. Question 7 options: represent various components of local tides that are resolved mathematically are predicted individually are added together to predict the height and timing of astronomical tides All of the above are correct. Only a and c are correct.
Complete Question
Partial tides __________.
Question 7 options:
a. represent various components of local tides that are resolved mathematically
b. are added together to predict the height and timing of astronomical tides
c. consist of 4 components due to the influence of celestial bodies
d. consist of up to 60 components due to astronomical and non-astronomical factors
e. All of the above except c are correct.
Answer:
Option E
Explanation:
Generally
Partial tides represent various components of local tides that are resolved mathematically
Partial tides are added together to predict the height and timing of astronomical tides
Partial tides consist of up to 60 components due to astronomical and non-astronomical factors
But Partial tides do not consist of 4 components due to the influence of celestial bodies
Therefore
All of the above except c are correct.
Option E
Is there any absolute rest or motion? Describe the types of motion with one example of each type
what are the limitation of clinical thermometer
Answer:
Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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Identify each action as a wave erosion war wind erosion
Answer:Lesson Objectives
Describe how the action of waves produces different shoreline features.
Discuss how areas of quiet water produce deposits of sand and sediment.
Discuss some of the structures humans build to help defend against wave erosion.
Vocabulary
arch
barrier island
beach
breakwater
groin
refraction
sea stack
sea wall
spit
wave-cut cliff
wave-cut platform
Introduction
Waves are important for building up and breaking down shorelines. Waves transport sand onto and off of beaches. They transport sand along beaches. Waves carve structures at the shore.
Wave Action and Erosion
All waves are energy traveling through some type of material, such as water (Figure below). Ocean waves form from wind blowing over the water.
Ocean waves are energy traveling through water.
The largest waves form when the wind is very strong, blows steadily for a long time, and blows over a long distance.
The wind could be strong, but if it gusts for just a short time, large waves won’t form. Wave energy does the work of erosion at the shore. Waves approach the shore at some angle so the inshore part of the wave reaches shallow water sooner than the part that is further out. The shallow part of the wave ‘feels’ the bottom first. This slows down the inshore part of the wave and makes the wave ‘bend.’ This bending is called refraction.
Wave refraction either concentrates wave energy or disperses it. In quiet water areas, such as bays, wave energy is dispersed, so sand is deposited. Areas that stick out into the water are eroded by the strong wave energy that concentrates its power on the wave-cut cliff (Figure below).
The wave erodes the bottom of the cliff, eventually causing the cliff to collapse.
Other features of wave erosion are pictured and named in Figure below. A wave-cut platform is the level area formed by wave erosion as the waves undercut a cliff. An arch is produced when waves erode through a cliff. When a sea arch collapses, the isolated towers of rocks that remain are known as sea stacks.
(a) The high ground is a large wave-cut platform formed from years of wave erosion. (b) A cliff eroded from two sides produces an arch. (c) The top of an arch erodes away, leaving behind a tall sea stack.
Wave Deposition
Rivers carry sediments from the land to the sea. If wave action is high, a delta will not form. Waves will spread the sediments along the coastline to create a beach (Figure below). Waves also erode sediments from cliffs and shorelines and transport them onto beaches.
Sand deposits in quiet areas along a shoreline to form a beach.
Beaches can be made of mineral grains, like quartz, rock fragments, and also pieces of shell or coral (Figure below).
Quartz, rock fragments, and shell make up the sand along a beach.
Waves continually move sand along the shore. Waves also move sand from the beaches on shore to bars of sand offshore as the seasons change. In the summer, waves have lower energy so they bring sand up onto the beach. In the winter, higher energy waves bring the sand back offshore.
Some of the features formed by wave-deposited sand are in Figure below. These features include barrier islands and spits. A spit is sand connected to land and extending into the water. A spit may hook to form a tombolo.
Examples of features formed by wave-deposited sand.
Shores that are relatively flat and gently sloping may be lined with long narrow barrier islands (Figure below). Most barrier islands are a few kilometers wide and tens of kilometers long.
(a) Barrier islands off of Alabama. A lagoon lies on the inland side. (b) Barrier islands, such as Padre Island off the coast of Texas, are made entirely of sand. (c) Barrier islands are some of the most urbanized areas of our coastlines, such as Miami Beach.
In its natural state, a barrier island acts as the first line of defense against storms such as hurricanes. When barrier islands are urbanized (Figure above), hurricanes damage houses and businesses rather than vegetated sandy areas in which sand can move. A large hurricane brings massive problems to the urbanized area.
Protecting Shorelines
Intact shore areas protect inland areas from storms that come off the ocean (Figure below).
Dunes and mangroves along Baja California protect the villages that are found inland.
Explanation:
Answer: Below
Explanation: Correct on Edmentum
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Cho hệ thống thùng lắc có mô hình tại vị trí đang xét như hình vẽ
Answer:
I can't understand this language!!!Answer:
vdhdbdnnsnsbdhhshzbhshsbbsbd is not ask you to be able and r in the exam qq and
The current in resistor Y is..?
(A)
Explanation:
We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that
[tex]P = VI[/tex]
Since resistor Y dissipates 100 W of power, we can solve for the current as
[tex]I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}[/tex]
The current in resistor Y is
a)1.0 AMichelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.
Answer:
The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.
Explanation:
Total Internal Reflection: A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface. The index of refraction for water is 1.33, and for the glass it is 1.43. What angle does the refracted ray in the water make with the normal
Answer:
θ₄ = 37.2º
Explanation:
For this exercise it must be solved in two parts, the first part we look for the critical angle, for this we use the law of refraction with the angle in the middle of transmission of tea = 90º
n₁ sin θ₁= n₂ sin 90
θ₁ = sin⁻¹ [tex]\frac{n_2}{n_1}[/tex]
θ₁ = sin⁻¹ (1.33 / 1.43)
θ₁ = 68.4º
They indicate that the angle of incidence is half of the critical angle
θ₃ = 68.4 / 2 = 34.2º
Let's use the law of refraction again
n₁ sin θ₃ = n₂ sin θ₄
sin θ₄ = [tex]\frac{n_1}{n_2}[/tex] sin θ₃
sin θ₄ = [tex]\frac{1.43}{1.33}[/tex] sin 34.2
θ₄ = sin⁻¹ 0.604345
θ₄ = 37.2º
11. From this lab we can conclude that a) the heat transferred when objects are rubbed together creates an energy that can cause objects to move towards or away from each other. b) objects such as balloons and sweaters have a natural affinity towards each other. They will attract each other whether they are rubbed together or not. c) charges exert forces on other charges. do) charges do not exert forces on other charges.
Answer:
c) charges exert forces on other charges.
Explanation:
When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.
It was recorded that the temperature of a body was 320 degree F determine the value of the temperature in kelvin
Answer:
433.15K
Explanation:
(320°F − 32) × 5/9 + 273.15 = 433.15K
