Answer:
The answer is D
ADvantage of friction
Answer:
1. Friction enables us to walk freely.
2. It helps to support ladder against wall.
3. It becomes possible to transfer one form of energy to another.
4. Objects can be piled up without slipping.
A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff the aircraft travels at 63.0 m/s, so that the air speed relative to the bottom of the wing is 63.0 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = [tex]\frac{\Delta F}{A \ \rho \ g} + v_1^2[/tex]
suppose that the area of the wing is A = 1 m²
we substitute
v₂² = [tex]\frac{1000}{1 \ 1.29 \ 9.8} + 63^2[/tex]
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s
a system absorb 500 J of heat and the same time 400J of work is done one the system find change in internal enery ?
Answer:
+ 900 J
Explanation:
Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,
ΔE = ΔU
ΔE = Q - W where Q = heat absorbed by system and W = work done by system
Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J
So, the values of the variables into the equation, we have
ΔE = Q - W
ΔE = + 500 J - (-400 J)
ΔE = + 500 J + 400 J
ΔE = + 900 J
So, the internal energy change, ΔE = + 900 J
define force and types of force
Answer:
Force is the strength or weight of things that depends on movement. The types of forces are conteact force, spring force, applied force, air resistance force, normal force, frictional force, tension force, and non-contact force.
Normal conversation has a sound level of about 60 dB. How many times more intense must a 10,000-Hz sound be compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness
Answer: A 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.
Explanation:
The formula used is as follows.
[tex]\beta = 10 dB log (\frac{I}{I_{o}})\\60 = 10 dB log (\frac{I}{I_{o}})[/tex]
[tex]I_{o} = 10^{-12}[/tex] normal threshold
The difference is sound level is as follows.
60 - 60 = 0
Hence,
[tex]0 = 10 dB [log (\frac{I_{f}}{I_{o}}) - log (\frac{I_{i}}{I_{o}})]\\log (\frac{1000}{I_{o}}) = log (\frac{10000 x}{I_{o}})\\log (10^{15}) = log (10^{16}x)\\15 = 16 + log x\\log x = 1\\x = 10[/tex]
This means that 10,000 Hz sound is 10 times more intense.
Thus, we can conclude that a 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.
A peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie plate moves through a distance of 208 in. Express the angular distance that the pie plate has moved through in revolutions, radians, and degrees.
Answer:
a) [tex]X_1=7.36rev[/tex]
b) [tex]X_2=46.22radians[/tex]
c) [tex]X_3=2649.6^o[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=9.00[/tex]
Distance [tex]x=208[/tex]
Generally the equation for circumference of a circle is mathematically given by
[tex]C=2 \pi r\\\\C=2*\pi*4.5[/tex]
[tex]C=28.3[/tex]
Therefore
Angular distance that the pie plate has moved through in revolutions is
[tex]X_1=\frac{x}{C}[/tex]
[tex]X_1=\frac{208}{28.3}[/tex]
[tex]X_1=7.36rev[/tex]
Generally Angular distance that the pie plate has moved through in radians is
[tex]X_2= 7.36rev* 2 \pi[/tex]
[tex]X_2=46.22radians[/tex]
Generally Angular distance that the pie plate has moved through in degrees is
[tex]X_3=7.36rev* 360[/tex]
[tex]X_3=2649.6^o[/tex]
A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: , where , , , and is not given. Which of the following represents a simplified expression for ?tanθ− agTo avoid making mistakes, the expression should not be simplified until the numerical values are substituted.gsinθ−agcosθThe single equation has two unknowns and cannot be solved with the information given.
Solution :
Given expression :
[tex]$\mu_k$[/tex]mgcosθ = mgsinθ − ma
Here, g = 9.8 [tex]m/s^2[/tex] , a = 3.60 [tex]m/s^2[/tex] , θ = 27°
Therefore,
[tex]$\mu_k mg \cos \theta = mg \sin \theta - ma$[/tex]
[tex]$\mu_k mg \cos \theta = m(g \sin \theta - a)$[/tex]
[tex]$\mu_k g \cos \theta = (g \sin \theta - a)$[/tex]
[tex]$\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$[/tex]
Mow calculating the coefficient of kinetic friction as follows :
[tex]$\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$[/tex]
[tex]$\mu_k=\frac{9.8 \times \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$[/tex]
[tex]$\mu_k=0.097$[/tex]
A person is driving a car down a straight road. The instantaneous acceleration is constant and in the direction of the car's motion. 1) The speed of the car is increasing. decreasing. constant. increasing but will eventually decrease. decreasing but will eventually increase.
Answer:Increasing
Explanation:
Given
Car is driven on the straight road with instantaneous acceleration in the direction of car's motion.
If instanateneous acceleration is constant then speed of car is increasing at a constant pace. As there are no turns on the road, therefore speed of car is increasing.
The speed of the car is "decreasing". A further description is provided in the below paragraph.
It's because the individual would be in a straightforward fashion. This same acceleration inclination comes contrary to the movement of the automobile. It indicates that it exerts pressure against the movement of the automobile. So, when it moves forward, the speed of the automobile decreases.
Thus the above answer is correct.
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a motor car reaches a velocity of 15m/s in 6s from rest on a perfect test track . what is the average acceleration
Answer:
[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]
PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
You sit with friends around a campfire, roasting marshmallows. Which
transfer of thermal energy involved in this system is an example of radiation
Answer:
The answer is c
Thermal energy moves within the air from the flames to the marshmallow.
Explanation:
Hope it helps
You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.
What is thermal Energy ?In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.
Hence option C is correct.
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What is the potential energy of a 7kg object 4m off the ground ?
please show your work
Answer:
Gravitational potential energy is mass of the object times the gravitational constant times the height of the object:
U = mgh (I will use 10 for the gravitational constant but you can use 9.8 or 9.81 or something even more accurate)
U = 280
The gravitational potential of the object is 280 joules
The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?
Answer:
Rahul's weight
Explanation:
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
Using formulas, Rahul's weight is equal to
W=mg
where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).
While a mason was working concrete into formwork, the formwork collapses. Who is BEST suites to rectify this problem? Mason Carpenter Project Manager O Construction Technician A device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop
Answer:
1. Carpenter
2. True
Explanation:
While a mason was working concrete into the formwork, the formwork collapses. The best person to rectify this problem is CARPENTER.
This is because it is the job of the Carpenter to design and build formwork, most especially wooden formwork. Formwork is like casing built to receive concrete and reinforcement during construction. Hence, when formwork collapses either due to stress, tension, or improper construction, it is the job of Carpenter to reconstruct the formwork or rectify the problem.
It is TRUE that when a device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop. However, this communication will be an instruction on what to do next, and it usually directs those responsible to take action where necessary. For example, a workshop manager communicates to a carpenter about the need to rectify a chair or table that has a defect.
You have a simple pendulum that oscillates with a period of 2 s as you stand on the surface of Earth. Your friend, an astronaut standing on the surface of the Moon, has a pendulum of the same length. What would be the period of oscillation of your friend’s pendulum?
a. Less than 2 s
b. The answer depends on whether the amplitudes are the same
c. More than 2 s
d. Exactly 2 s
Answer:
c. More than 2 s
Explanation:
First, we will find the length of the pendulum:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\2\ s = 2\pi \sqrt{\frac{l}{9.81\ m/s^2}}\\\\4\ s^2 = 4\pi^2 (\frac{l}{9.81\ m/s^2})\\\\l = \frac{(4\ s^2)(9.81\ m/s^2)}{4\pi^2} \\\\l = 0.99\ m[/tex]
Now, the value of g becomes 1.625 m/s² on the surface of the moon. So the time period will be:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\T = 2\pi \sqrt{\frac{0.99\ m}{1.625\ m/s^2}}\\\\[/tex]
T = 4.9 s
Therefore, the correct option is:
c. More than 2 s
Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.
Explanation:
Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.
The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.
The relationship between Gibbs free energy change and reaction quotient of the reaction is:
[tex]\Delta G=\Delta G^o+RT ln Q[/tex] ......(1)
where,
[tex]\Delta G[/tex] = Gibbs free energy change
[tex]\Delta G^o[/tex] = Standard Gibbs free energy change
R = Gas constant
T = Temperature
At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:
[tex]\Delta G^o=-RT ln Q[/tex] ...(2)
Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks. Assume the dome has a diameter of 25.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 10^6 V/m.
Required:
a. What is the maximum potential of the dome?
b. What is the maximum charge on the dome?
Answer:
(a) V = 3.75 x 10^5 V
(b) q = 5.2 x 10^-6 C
Explanation:
Diameter, d = 25 cm
radius, r = 12.5 cm = 0.125 m
Electric field, E = 3 x 10^6 V/m
(a) The maximum potential is given by
[tex]V = E \times r \\\\V = 3\times 10^6\times 0.125\\\\V = 3.75\times10^5 V[/tex]
(b) The charge is given by
[tex]V = \frac{k q}{r}\\\\3.75\times10^5=\frac{9\times10^9\times q}{0.125}\\\\q = 5.2\times 10^{-6} C[/tex]
How is wind generated?
O A. Air molecules move from areas of low pressure to areas of high
pressure.
O B. Air molecules move more slowly where the temperature is higher
and the pressure is lower.
C. Air molecules move more quickly where the temperature is lower
and the pressure is higher.
D. Air molecules move from areas of high pressure to areas of low
pressure.
Answer:
Explanation:
Wind is caused by the uneven heating of the atmosphere by the sun, variations in the earth's surface, and rotation of the earth. ... Wind turbines convert the energy in wind to electricity by rotating propeller-like blades around a rotor. The rotor turns the drive shaft, which turns an electric generator
HELP ME PLS!!!!
Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge
Answer:
the answer is c which is a+2 charge
Explanation:
Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.
The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.
What are cations and anions?In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.
This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.
Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.
The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.
Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.
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A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,
increasing the speed to 2 m/s. How much work did the man do?
A. 640 J
B. 360 J
C. 1360 J
D. 1000 J
Work done by man will be A. 640 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work energy theorem
Work done = final Kinetic energy - initial kinetic energy
= KE (final) - KE (initial )
= 1/2 m ([tex]v^{2}[/tex]) - 1/2 m ([tex]u^{2}[/tex])
= 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])
= 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])
= 250 * 2.56 = 640 J
correct answer is A. 640 J
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3. What is electric current?
The flow of moving electrons
electrons that move one time
Answer:
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.
Answer:
The flow of moving electrons
Calculate the forces that the supports \rm A and \rm B exert on the diving board shown in when a 58-\rm kg person stands at its tip.
Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is given by g = Gm/R2 . What is the acceleration of a satellite moving in a circular orbit round the earth of radius 2R
Explanation:
The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.
[tex]W=F_G[/tex]
[tex]mg = G \dfrac{mM}{R^2}[/tex]
which gives us an expression for the acceleration due to gravity g as
[tex]g = G\dfrac{M}{R^2}[/tex]
At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is
[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]
Simplifying this, we get
[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
What is meant by kinetic friction ?Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.
Here,
Mass of the box, m = 150 lb = 68.1 kg
Coefficient of kinetic friction, μ = 0.45
Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,
F(k) = μmg
F(k) = 0.45 x 68.1 x 9.8
F(k) = 300.32 N
Now, the box sits on a ramp inclined at 60°
Coefficient of kinetic friction, μ = 0.45
The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.
So,
Frictional force, F(k)' = μmgcosθ
F(k)' = 0.45 x M x 9.8 x cos 60
F(k)' = 2.2M
Weight of the box acting horizontally,
W = Mgsinθ
W = M x 9.8 x sin60
W = 8.5M
Therefore, net force,
Fn = W - F(k)'
Fn = 8.5M - 2.2M
Fn = 6.3M
The total force acting on the box is
F = F(k) - Fn
ma = 300.32 - 6.3M
Since, the box is moving with constant speed, the acceleration, a = 0
Therefore,
300.32 - 6.3M = 0
6.3M = 300.32
M = 300.32/6.3
M = 47.7 kg = 105.16 pound
Hence,
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
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Identify the reactants in the combustion of methane: CH4 + O2 CO2 + O°H
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
Air is classified as a mixture. Option D is the correct answer.
Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.
In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.
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Question 18/55 (2 p.)
A vibrating object produces ripples on the surface of a liquid. The object completes 20 vibrations
every second. The spacing of the ripples, from one crest to the next, is 3.0 cm.
What is the speed of the ripples?
D
C 60 cm/s
120 cm/s
A 0.15cm/s
B 6.7 cm/s
Answer:
the correct answer is C v = 60 cm / s
Explanation:
The speed of a wave is related to the frequency and the wavelength
v = λ f
They indicate that the object performs 20 oscillations every second, this is the frequency
f = 20 Hz
the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength
λ = 3 cm = 0.03 m
let's calculate
v = 20 0.03
v = 0.6 m / s
v = 60 cm / s
the correct answer is C
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of 33 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
1. What is the ratio of the electric force on the bee to the bee's weight?
2. What electric field strength would allow the bee to hang suspended in the air?
3. What electric field direction would allow the bee to hang suspended in the air?
Answer:
A) 3.367 × 10^(-6)
B) 2.97 × 10^(7) N/C
C) Upwards
Explanation:
We are given;
Mass of bee; m = 100 mg = 100 × 10^(-6) kg
Charge on bee;q=33 pC = 33 × 10^(-12)C
Electric field strength; E = 100 N/C
A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N
Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N
ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)
B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;
mg = qE
100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E
E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))
E = 2.97 × 10^(7) N/C
C) From Newton's law, sum of forces = 0.
Thus;
F_n + F + W = 0
Where F is the normal force.
Thus;
F_n = -(F + W)
F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))
F_n = -9.8 × 10^(-4) N
Thus, applied electric field is;
E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C
This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.
The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant ( ε=24.78x10-12 F/m),the capacitance of nerve cell is
Answer:
9.965 nF
Explanation:
The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m
So, C = εA/d
C = ε2πrL/d
Substituting the of the values variables into the equation, we have
C = ε2πrL/d
C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m
C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m
C = 996463 × 10⁻¹⁴ F
C = 9.96463 × 10⁻⁹ F
C = 9.96463 nF
C ≅ 9.965 nF
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
