Complete Question
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .
Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Part B
If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
Answer:
Part A
[tex]R = 1081 \ J[/tex]
Part B
[tex]h = 169.7 \ m[/tex]
Explanation:
From the question we are told that
The mass of the blade is [tex]m_b = 0.65 \ kg[/tex]
The length is [tex]l = 0.55 \ m[/tex]
The angular speed is [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]
Generally the moment of inertia of the of this mower is mathematically evaluated as
[tex]I = \frac{m_b * l^2 }{12}[/tex]
substituting values
[tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]
[tex]I = 0.016 \ kg m^2[/tex]
Generally the rotational kinetic energy of the bland is
[tex]R = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]
[tex]R = 1081 \ J[/tex]
At point where the gravitational potential energy is equal to the rotational kinetic energy we have that
[tex]P = R = m_b * h * g[/tex]
Where P is the gravitational potential energy
substituting values
[tex]1081 = 0.65 * 9.8 * h[/tex]
=> [tex]h = 169.7 \ m[/tex]
What is the frequency if 140 waves pass in 2 minutes?
Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
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?
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.
Nuclear fusion in our Sun happens when
- hydrogen atoms combine to make helium atoms and release energy
- uranium atoms break apart and release energy
- hydrogen atoms are burned and release energy
- helium atoms break apart and release energy
Answer:
A
Explanation:
Fussion occurs when elements of lower atomic mass combines to form that of a larger atomic mass, releasing energy in the process .
Hydrogen has a lower atomic mass than Helium.
A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 990 kg ? Neglect the buoyant force on the cargo volume itself. Assume gases are at 0∘C and 1 atm pressure (rhoair = 1.29 kg/m3, rhohelium = 0.179 kg/m3).
Answer:
The mass of the cargo is [tex]M = 188.43 \ kg[/tex]
Explanation:
From the question we are told that
The radius of the spherical balloon is [tex]r = 7.40 \ m[/tex]
The mass of the balloon is [tex]m = 990\ kg[/tex]
The volume of the spherical balloon is mathematically represented as
[tex]V = \frac{4}{3} * \pi r^3[/tex]
substituting values
[tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]
[tex]V = 1697.6 \ m^3[/tex]
The total mass the balloon can lift is mathematically represented as
[tex]m = V (\rho_h - \rho_a)[/tex]
where [tex]\rho_h[/tex] is the density of helium with a value of
[tex]\rho_h = 0.179 \ kg /m^3[/tex]
and [tex]\rho_a[/tex] is the density of air with a value of
[tex]\rho_ a = 1.29 \ kg / m^3[/tex]
substituting values
[tex]m = 1697.6 ( 1.29 - 0.179)[/tex]
[tex]m = 1886.0 \ kg[/tex]
Now the mass of the cargo is mathematically evaluated as
[tex]M = 1886.0 - 1697.6[/tex]
[tex]M = 188.43 \ kg[/tex]
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 :)
Someone please helppppppp!!!!!
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
The air flowing over the top of the wing travels
in the same amount of time than the air
flowing beneath the wing.
Answer: Short Answer: NO ( In Most Cases)
Explanation:
If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
how does the statement " silence is golden " relate to ethics in communicating at the workplace.?
Answer:
Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.
The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.
A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to
Answer:
The time when the ball strikes the ground is closest to [tex]t_t = 9.4 \ s[/tex]
Explanation:
From the question we are told that
The time of projection is t = 0.0 s
The distance of the point from the ground is [tex]d = 90 \ m[/tex]
The initial velocity of the ball is [tex]v _i = 36 .2 \ m/s[/tex]
generally the time required to reach maximum height is
[tex]t_r = \frac{g}{v}[/tex]
Where is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
Substituting values
[tex]t_r = \frac{36.2}{9.8}[/tex]
[tex]t_r = 3.69 s[/tex]
when returning the time and velocity at the roof level is t = 3.69 s and u = 36.2 m/s this due to the fact that air resistance is negligible
The final velocity at which it hit the ground is
[tex]v_f^2 = u^2 + 2ag[/tex]
So
[tex]v_f = \sqrt{ u^2 + 2gs}[/tex]
substituting values
[tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]
[tex]v_f = 55.45 \ m/s[/tex]
The time taken for the ball to move from the roof level to the ground is
[tex]t_g = \frac{v-u}{a}[/tex]
substituting values
[tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]
[tex]t_g = 1.96 \ s[/tex]
The total time for this travel is
[tex]t_t = t_g + 2 t_r[/tex]
[tex]t_t = 1.96 + 2(3.69)[/tex]
[tex]t_t = 9.4 \ s[/tex]
Michelson and Morley's experiment is widely considered to have been:______
a. a success because it detected a shift in the interference pattern.
b. a failure because it detected a shift in the interference pattern.
c. a success because it did not detect a shift in the interference pattern.
d. a failure because it did not detect a shift in the interference pattern.
e. lacking the necessary precision to determine a shift in the interference pattern.
Answer:
The correct answer is option (c) a success because it did not detect a shift in the interference pattern.
Explanation:
In Michelson and Morley experiment it was considered to be successful.
They both found out that the experiment that was carried out was not a failure since it did not detect any shift in the interference pattern.
With this findings it was widely regarded as correct and precise.
A projectile is defined as
Answer:
By definition, a projectile has a single force that acts upon it - the force of gravity.
Explanation:
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.
// have a great day //
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
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Answer:
0.19rad/s and Yes
Explanation:
From the principle of conservation of momentum it means momentum before and after collision is the same.
Momentum before collision is 0.700 kg×12 = 8.4Ns
Momentum of the door = mass of door × velocity of door
8.4Ns = mass of door × velocity of door
Velocity of door = 8.4Ns/45 =0.19m/s
But velocity V= w×r ;
w-angular velocity
r- raduis = width
w= 0.19/1m = 0.19rad/s
2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
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
Which statement BEST explains the relationship between voltage, current, and power?
A. If voltage increases and everything else remains constant, then power will increase.
B. If voltage increases and everything else remains constant, then power will decrease.
C. If current decreases and everything else remains constant, then power will increase.
D. Voltage and power are inversely related.
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
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
[tex]L_{f}[/tex] = I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
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]
Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=
Answer:
Explanation shown below.
Explanation:
1.The number of images formed by 2 parallel mirrors is an infinite number of images.
2. The characteristics of a plane mirror is such that the object distance equals the image distance.
Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.
Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.
Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .
Hence the shortest distances are 8.7cm and 68.9cm
3. The number of reflections is infinite for both cases.
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
The Nardo ring is a circular test track for cars. It has a circumference of 12.5km. Cars travel around the track at a constant speed of 100km/h. A car starts at the easternmost point of the ring and drives for 15 minutes at this speed.
1. What distance, in km, does the car travel?
2. What is the magnitude of the car's displacement, in km, from its initial position?
3. What is the speed of the car in m/s?
Answer:
1. 25 Km
2. zero
3. 27.7 m/s
Explanation:
Data provided in the question:
Circumference of the track = 12.5 km
Speed of the car = 100 Km/h
Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr
Now,
1. Distance traveled = Speed × Time
= 100 × [tex]\frac{15}{60}[/tex]
= 25 Km
2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)
Which means that the car is again at the initial position
Therefore, The displacement is zero.
3. Speed of car in Km/hr = 100 Km/h
now,
1 Km = 1000 m
1 hr = 3600 seconds
therefore,
100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s
= 27.7 m/s
Hence, the speed of car in m/s = 27.7
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:
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
Someone please helpp me out thanks !
Answer:
Silver.
Explanation:
To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:
Mass of metal (m) = 18.15g
Length (L)= 1.2cm
Volume (V) = L³ = 1.2³ = 1.728cm³
Density =.?
The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:
Density = Mass /volume
With the above formula, we can obtain the density of the metal as follow:
Mass = 18.15g
Volume = 1.728cm³
Density =.?
Density = Mass /volume
Density = 18.15g/1.728cm³
Density of the metal = 10.50g/cm³
Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.
Therefore, the metal is silver.
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
