Answer:
Only a decreasing gravitational force that acts downward
Explanation:
The gravitational force is the gravitational pull which attract a mass of smaller size by the mass of a bigger size. It is the force which attract two masses close to each other.
In the context, when a ball is tossed up from the surface of an asteroid that have no atmosphere, the ball rises up and then falls back to the surface of the asteroid. The ball falls back because the gravitational pull of the asteroid pulls back the ball to its surface. Thus a decreasing gravitational force acts on the ball in the downward direction while the ball is in its way up.
what is the meaning of friend ?
Answer:
person that you know and like (not a member of your family), and who likes you
Seawater fills a tank to a depth of 12 ft. If the specific gravity of seawater is 1.03 and the atmospheric pressure at this location is 14.8 psi, the absolute pressure (psi) at the bottom of the tank is most nearly:______.
A. 5.4
B. 20.2
C. 26.8
D. 27.2
E. 10.8
Answer: (b)
Explanation:
Given
Depth of tank is [tex]h=12\ ft\ \text{or}\ 3.65\ m[/tex]
Specific gravity of seawater is [tex]S.G.=1.03[/tex]
Pressure difference due to column of water is
[tex]\Rightarrow \Delta P=\rho gh\\\Rightarrow \Delta P=1.03\times 10^3\times 9.8\times 3.65\\\Rightarrow \Delta P=36.84\times 10^3\ Pa\ \text{or}\\\Rightarrow \Delta P=5.34\ psi[/tex]
So, absolute pressure is given by
[tex]\Rightarrow P_{abs}=P_{atm}+\Delta P\\\Rightarrow P_{abs}=14.8+5.34\\\Rightarrow P_{abs}=20.14\approx 20.2\ psi[/tex]
Thus, option (b) is correct.
Visible matter belonging to the Milky Way Galaxy can be traced out to about 50,000 light years from the center.
a. True
b. False
Answer:
b. False
Explanation:
The visible matter that belongs to the Milky way Galaxy are traced out to be about 50 kpc distance from the center.
Kpc stands for kiloparsec. It is the unit of measurement of distance.
A parsec is[tex]$\text{ used to measure large distances}$[/tex] of the astronomical objects that lies [tex]$\text{outside the solar system}$[/tex], mainly where galaxies are involved.
1 kiloparsec is 1000 parsec and is equal to 3260 light years.
So the visible matter is about 163,078 light years away.
Hence the answer is FALSE.
PLEASE HELP!!!!
A person pushes attempts to push a couch with a 25 N force. The couch, however, doesn't move. What is the static friction force acting on the couch? *
A. 25 N
B. 0 N
C. 50 N
D. There is no static friction the ONLY force acting on the couch is the push
Static friction cancels out the force of the push, so it also has a magnitude of 25 N.
A ball is dropped out of a window and hits the ground at 14.5 m/s. How long did it take to fall to the ground?
Answer:
Explanation:
Use the one-dimensional equation:
[tex]v_f=v_0+at[/tex] which says that the final velocity of a falling object is equal to its initial velocity times the acceleration of gravity times the time it takes to fall. We have the final velocity, -14.5 (negative because its direction is down and down is negative), initial velocity is 0 (because it was held still by someone before it was dropped), and acceleration is -9.8 (negative again, because direction is down while acceleration increases). Filling in:
-14.5 = 0 - 9.8t and
-14.5 = -9.8t so
t = 1.5 seconds
If a car's velocity is 30 m/s and it drives at this velocity for 4 seconds, how far did it go?
Answer:
120 m
General Formulas and Concepts:
Kinematics
VelocityDisplacementDistanceTimeExplanation:
Step 1: Define
Identify
[Given] v = 30 m/s
[Given] t = 4 s
Step 2: Solve
Multiply [Cancel out units]: 30 m/s · 4 s = 120 mAnswer:
[tex]\boxed {\boxed {\sf 120 \ meters}}[/tex]
Explanation:
Distance, or how far an object travels, is the product of velocity and time.
[tex]d= v*t[/tex]
The velocity is 30 meters per second and the time is 4 seconds.
v= 30 m/s t=4 sSubstitute the values into the formula.
[tex]d= 30 \ m/s * 4 \ s[/tex]
Multiply. The units of seconds (s) cancel.
[tex]d= 30 \ m * 4[/tex]
[tex]d=120 \ m[/tex]
The car travels a distance of 120 meters in 4 seconds at a velocity of 30 meters per second.
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
Answer:
Option 2
Explanation:
The complete question is
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
TopBottomLeftRightSolution
It is the bottom of the bucket that will high pressure because of the additional weight of 50 Kg boy along with the weight of the water and the tub itself.
Pressure acts in the down ward direction and is equal to the force/weight divided by the area.
Hence, option 2 is correct
A single-story retail store wishes to supply all its lighting requirement with batteries charged by photovoltaic cells. The PV cells will be mounted on the horizontal rooftop. The time-averaged lighting requirement is 10 W/m2 , the annual average solar irradiance is 150 W/m2 , the PV efficiency is 10%, and the battery charging/discharging efficiency is 80%. What percentage of the roof area will the PV cells occupy
Answer:
83.33% of the roof area will be occupied by the PV cells
Explanation:
Given the data in the question;
time-averaged lighting requirement [tex]P_{lighting[/tex] = 10 W/m²
the annual average solar irradiance [tex]q_{solar[/tex] = 150 W/m²
the PV efficiency η[tex]_{pv[/tex] = 10% = 0.1
battery charging/discharging efficiency η[tex]_{battery[/tex] = 80% = 0.8
we know that; Annual average power to the light = [tex]P_{lighting[/tex] × A[tex]_{roof[/tex]
Now, the electrical power delivered by the solar cell battery system will be;
⇒ [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
[tex]P_{lighting[/tex]A[tex]_{roof[/tex] = [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
Such that;
A[tex]_{pv[/tex] = [tex]P_{lighting[/tex]A[tex]_{roof[/tex] / [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = [tex]P_{lighting[/tex] / [tex]q_{solar[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
so we substitute
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / 12 W/m²
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 0.8333
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = (0.8333 × 100)%
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 83.33%
Therefore, 83.33% of the roof area will be occupied by the PV cells.
A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
ocean floor and is received 15s later. What is the distance between the submarine and the ocean floor?
Answer:
the distance between the submarine and the ocean floor is 11,250 m
Explanation:
Given;
speed of the wave, v = 1500 m/s
time of motion of the wave, t = 15 s
The time taken to receive the echo is calculated as;
[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]
Therefore, the distance between the submarine and the ocean floor is 11,250 m
4) The SI unit of time is second. why?
Answer:
Second, fundamental unit of time, now defined in terms of the radiation frequency at which atoms of the element cesium change from one state to another. The second was formerly defined as 1/86,400 of the mean solar day—i.e., the average period of rotation of the Earth on its axis relative to the Sun.
source britannica
Explanation:
Is a measurement is precuse it must also be accurate
Two trains are moving at 50 m/s in opposite directions on the same track. The engineers see simultaneously that they are on a collision course and apply the brakes when they are 1,100 m apart. Assuming both trains have the same acceleration, what must this acceleration be (in m/s2) if the trains are to stop just short of colliding
Answer:
Acceleration is 2.5 m/s^2
Explanation:
Let train 1 travel x meters and train 2 travels 1000-x meters
As per the las of acceleration
[tex]v^2 = u^2 + 2ax[/tex]
Substituting the given values, we get -
[tex]0 = 50^2 + 2ax\\ax = -1250\\[/tex]---Eq (1)
Similarly
[tex]0 = 50^2 + 2a(1000-x)\\a(1000-x) = 1250[/tex]-- Eq (2)
Substituting the value of ax from eq (1) into Eq(2), we get -
1000*a - 1250 = 1250
a = 2500/1000 = 2.5 m/s^2
Many adventures like to go rafting on the Colorado River through Grand Canyon National Park. There are many locations where the river becomes more narrow, both the distance between the canyon walls as well as the depth changes due to debris like boulders on the bottom of the river; this leads to changes in the water speed. In the park, the Colorado River has an average width of 100m and an average depth of 8m, and an average speed of 3 m/s. At the Lava Falls Rapids, the river has an average width of about 25m and an average depth of about 15m. What is the approximate speed of the water in this location
Answer:
6.4 m/s
Explanation:
Given that :
The average width of the Colorado river = 100 m
Average depth of the river is = 8 m
Therefore, area = [tex]$A_1= 100 \ m \times 8 \ m$[/tex]
Speed of the river, [tex]$v_1 = 3 \ m/s$[/tex]
After the lava falls on the river,
Width of the river becomes = 25 m
Depth of the river became = 15 m
Therefore, area = [tex]$A_2= 25 \ m \times 15 \ m$[/tex]
Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,
[tex]$Q_1=Q_2$[/tex]
[tex]$A_1v_1=A_2v_2$[/tex]
∴ [tex]$100 \times 8 \times3 = 25 \times 15 \times v_2$[/tex]
[tex]$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$[/tex]
= 6.4 m/s
Therefore, the speed of the river in this location is 6.4 m/s
allung d Uall, Wily
In which condition is mirage seen ? Why is light dispersed ?
Answer:
Mirage is a phenomenon which can be seen when the surface air gets heated up and it becomes lighter. Lighter air moves up in the atmoshphere.
Explanation:
When the lighter air from cooler areas to warmer areas are refracted and they bent upwards.and it dispers
,
How long would it take a 4,560 watt motor to raise a 166 kg piano to an apartment window
15 meters above the ground?
Answer:
Explanation:
We need the power equation here:
P = W/t where W is work and is defined as
W = F*displacement.
Force is a measure in Newtons, which is also weight. We have the mass of the piano, but we need to find the weight:
w = mg so
w = 166(9.8) so
w = 1600N, rounded to the correct number of sig dig. We use that now in the power equation:
[tex]4560=\frac{(1600)(15)}{t}[/tex] and isolating the unknown:
[tex]t=\frac{(1600)(15)}{4560}[/tex] so
t = 5.3 seconds
what happened in my room
Answer:
A GHOST CAME! Booooo!!!!!!Hah lol
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
Required:
a. What is the charge on the inner surface of the shell?
b. What is the charge on the outer surface of the shell?
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Reasons for change of charge on a body
Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.
While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Learn more about charge here: https://brainly.com/question/18102056
Convert 15000kg/m cube
into gm/cm cube
please write the process also
Answer:
15000 Kilograms/Cubic Meters (kg/m3) = 15 Grams/Cubic Centimeters (g/cm3)
Explanation:
1 g/cm3 is equal to 1000 kilogram/cubic meter. To convert 100 gram into kg then divide it by 1000 i.e. 100/1000 = 0.1 kg. To convert any value of gm/cm3 into kg/m3 then multiply it by 1000.
15000 kg / m^3 =
15000 × 10^3 g / m^3 =
15000 × 10^3 × 10^3 mg / m^3 =
15 × 10^9 mg / m^3 =
15 × 10^9 × 10^(-3) mg / dm^3 =
15 × 10^9 × 10^(-3) × 10^(-3) mg / cm^3 =
15 × 10^9 × 10^(-6) mg / cm^3 =
15 × 10^( 9 - 6 ) mg / cm^3 =
15 × 10^3 mg / cm^3 =
15000 mg / cm^3 =
Look : We found the exact thing we had ...
WoW ...
We got a point ;
Remember from now on :
kg / m^3 = mg / cm^3
A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance
Answer:
[tex]R=2.78\ \Omega[/tex]
Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :
[tex]P=I^2R[/tex]
Put all the values to find R.
[tex]R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega[/tex]
So, the resistance is equal to [tex]2.78\ \Omega[/tex].
a. Green light shines through a 100mm-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.
b. Green light shines through a 100-μ m-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.
Answer:
a) size of the bright spot is proportional to the hollow size
b) as the size of the hole increases, the circular point decreases.
Explanation:
a) In this case the diameter of the hole is much greater than the wavelength, as the size of the hole is many orders greater than the wavelength we are in the part of geometric optics,
Consequently the size of the bright spot is proportional to the hollow size.
Consequently the size increases
b) in this case the hole diameter d = 100 10⁻⁶m and the wavelength that for the green color is lam = 500 nm = 5 10⁻⁷ m
We see that angles are very small so the wavelength of the office is greater than the wavelength, but you can observe the effects of diffraction
d sin θ = 1.22 m λ
the numerical constant appears by solving the equation in polar coorθdinates, because the hole is circular
the first zero occurs for m = 1
sin θ = 1.22 λ / d
In these experiments the angles are small
sin θ = θ
we substitute
θ = 1.22 λ/ d
θ = 1.22 500 10⁻⁹ / 100 10⁻⁶
θ = 6.1 10⁻³
without the hole diameter increases by 20%
d’ = 1.2 d
we substitute
θ'= 1.22 λ / d'
θ’ = 1.22 λ /1.2 d
θ‘= 1.22 λ /d [tex]\frac{1}{1.22}[/tex]
θ ’= θ 0.83
θ ’= 6.1 10⁻³ 0.83
θ' = 5 10⁻³ rad
Therefore, the answer is that as the size of the hole increases, the circular point decreases.
A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact
Answer:
The speed of the bullet just before the impact is 701 m/s
Explanation:
Given;
mass of the bullet, m₁ = 20 g = 0.02 kg
mass of the block, m₂ = 2.8 kg
displacement of the block, d = 9 mm = 9 x 10⁻³ m
duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s
Apply the principle of conservation of energy;
The final kinetic energy of the bullet = maximum potential energy of the block
[tex]\frac{1}{2} m_1v^2 = m_2gh\\\\v^2 = \frac{2m_2gh}{m_1} \\\\v= \sqrt{\frac{2m_2gh}{m_1} } \\\\v = \sqrt{\frac{2 \times 2.8 \times 9.8 \times (9\times 10^{-3})}{0.02} } \\\\v = 4.97 \ m/s[/tex]
Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₁ is the initial velocity of the bullet
u₂ is the initial velocity of the block = 0
m₁u₁ + 0 = v(m₁ + m₂)
m₁u₁ = v(m₁ + m₂)
0.02u₁ = 4.97(2.8 + 0.02)
0.02u₁ = 14.02
u₁ = 14.02 / 0.02
u₁ = 701 m/s
Therefore, the speed of the bullet just before the impact is 701 m/s
John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. The rope makes an angle of 25 with the ground He is pulling horizontally with a constant force of 400 N. John and manages to get the sled going from 0 to 4 m/s in 5 s. The force due to friction on the sled is 310 N. What is the mass of Laura and the sled combined
which of the following is not a good working habit in doing an embroidery
Answer:
Where is following ??
Post proper Ques
Fill in the blanks. Power station produce electricity at __________.
Answer:
heat
Explanation:
heat is the answer hdhhdhdhdhdhdhdhdhdhd
Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion with a negative velocity and a negative acceleration.
Answer:
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
Explanation:
In a vertical movement the acceleration is always downwards, therefore negative since it is created by the attraction of the Earth on the body.
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
A student removes a 10.5 kg stereo amplifier from a shelf that is 1.82 m high. The amplifier is lowered at a constant speed to a height of 0.75 m. What is the work done by (a) the person and (b) the gravitational force that acts on the amplifier
Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J
How do a parachutes work??4-5 sentences plsss help rn
Answer:
Explanation:
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.
C.
A palm fruit dropped to the ground from the top of
a tree 45m tall. How long does it take to reach the
ground? A. 9s B. 4.5s C. 6 D. 7.5s E. 35
(g = 10ms2).
Answer:
b 4.5
Explanation:
time=distance/speed
magnetism/ magnetic field ana magnetic forces
Answer:
Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.
A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant
(a) 4.20 J
(b) 16.74 J
Explanation:For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J