Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec[/tex]
Velocity of the string wave is equal to [tex]v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec[/tex]
Power of wave propagation is equal to [tex]P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt[/tex]
So power of the wave will be equal to 5.464 watt
a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a voltage of 20.0 V and has air betweeen the plates. How much work is done on the capacitor as the plate seperation is inceased to 2.00 cm?
Answer:
[tex]W = -2.76\times 10^{-9}~J[/tex]
Explanation:
The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.
The potential energy is given by the following formula:
[tex]U = \frac{1}{2}CV^2[/tex]
where C can be calculated using the plate separation and area.
[tex]C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon[/tex]
Therefore, the potential energy in the first case is
[tex]U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon[/tex]
In the second case:
[tex]C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon[/tex]
The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.
So, the difference in the potential energy is
[tex]W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J[/tex]
A series RL circuit with L = 3.00 H and a series RC circuit with C = 3.00 F have equal time constants. If the two circuits contain the same resistances R, (a) what is the value of R and (b) what sit the time constant?
Answer:
(a) R = 1Ω
(b) τ = 3
Explanation:
The time constants of the given circuits are as follows
[tex]\tau_{RL} = \frac{L}{R}\\\tau_{RC} = RC[/tex]
If the two circuits have equal time constants, then
[tex]\frac{L}{R} = RC\\R^2 = \frac{L}{C} = \frac{3}{3} = 1\\R = 1\Omega[/tex]
Therefore, the time constant in any of the circuits is
[tex]\tau = RC = 3[/tex]
(a). Value of resistance R is 1 ohm.
(b). The Value of time constant will be 3 second.
The time constant is defined as, time taken by the system to reach at 63.2% of its final value.
Time constant in RL circuit is, [tex]=\frac{L}{R}[/tex]
Time constant in RC circuit is, [tex]=RC[/tex]
Since, in question given that both circuit have same time constant.
So, [tex]RC=\frac{L}{R}\\\\R^{2}=\frac{L}{C}\\\\R=\sqrt{\frac{L}{C} }[/tex]
substituting L = 3H and C = 3F in above expression.
[tex]R=\sqrt{\frac{3}{3} }=1ohm[/tex]
Time constant = [tex]RC=1*3=3s[/tex]
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If the wind velocity is 43 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 212 km/h
Answer:
11.48 degree N of W
Explanation:
We are given that
Wind velocity=[tex]v_w=-43[/tex]km/h
Because wind is blowing towards south
Air speed=[tex]v_a=-212km/h[/tex]
Because the captain want to move with air speed in west direction.
x component of relative velocity=-212 km/h
y-Component of relative velocity=-43km/h
Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
[tex]\theta=tan^{-1}(\frac{-43}{-212}=11.48^{\circ}[/tex]N of W
Hence, the direction in which the pilot should set her course to travel due west=11.48 degree N of W
At 47 °C, what is the fraction of collisions with energy equal or greater than an activation energy of 88.20 kJ/mol?
Answer:
The fraction of collision is [tex]4.00\times10^{-15}[/tex]
Explanation:
Given that,
Temperature = 47°C
Activation energy = 88.20 KJ/mol
From Arrhenius equation,
[tex]k=Ae^{-\dfrac{E_{a}}{RT}}[/tex]
Here, [tex]e^{-\dfrac{E_{a}}{RT}}[/tex]=fraction of collision
We need to calculate the fraction of collisions
Using formula of fraction of collisions
[tex]f=e^{-\dfrac{E_{a}}{RT}}[/tex]
Where f = fraction of collision
E = activation energy
R = gas constant
T = temperature
Put the value into the formula
[tex]f=e^{-\dfrac{88.20}{8.314\times10^{-3}\times(47+273)}}[/tex]
[tex]f=4.00\times10^{-15}[/tex]
Hence, The fraction of collision is [tex]4.00\times10^{-15}[/tex]
How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from the horizontal?
Answer:
W = 7591.56 J
Explanation:
given,
distance of the box, d = 37 m
Force for pulling the box, F = 217 N
angle of inclination with horizontal,θ = 19°
We know,
Work done is equal to product of force and the displacement.
W = F.d cos θ
W = 217 x 37 x cos 19°
W = 7591.56 J
Hence, the work done to pull the box is equal to W = 7591.56 J
Final answer:
The work done to slide the box is 7586.09 Joules.
Explanation:
To calculate the work done to slide a box along the ground, we can use the formula:
Work = Force x Distance x cos(theta)
Where:
Force = 217 N (the force applied to pull the box)
Distance = 37 meters (the distance the box is being slid)
theta = 19° (the angle between the applied force and the horizontal)
Plugging in these values into the formula, we get:
Work = 217 N x 37 m x cos(19°)
Calculating this using a calculator, we find that the work done to slide the box is approximately 7586.09 Joules.
A space-based telescope can achieve a diffraction-limited angular resolution of 0.05″ for red light (wavelength 700 nm). What would the resolution of the instrument be (a) in the infrared, at 3.5 µm, and (b) in the ultraviolet, at 140 nm?
Answer:
a) [tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
b) [tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
Explanation:
We are comparing two wavelengths with the radius and diameter constant, and if we want to compare it, we need to use the following formula:
[tex]\frac{\theta_1}{\theta_2}= \frac{\lambda_1}{\lambda_2}[/tex]
Where [tex] \theta[/tex] represent the angular resolution and [tex]\lambda[/tex] the wavelength.
So if we have a fixed resolution and wavelength 1 and we want to find the resolution for a new condition we can solve for [tex] \theta_2[/tex] and we got
[tex] \theta_2 = \theta_1 \frac{\lambda_2}{\lambda_1}[/tex]
Part a
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm *\frac{1 \mu m}{1000 nm} = 0.7 \mu m[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the infrared case we know that [tex] \lambda_2 = 3.5 \mu m[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
Part b
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the ultraviolet case we know that [tex] \lambda_2 = 140 nm[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Explanation:
A.
Given:
V = 13 m/s
t = 4 s
Constant acceleration, a= (V-Vi)/t
= 13/4
= 3.25 m/s^2
F = mass * acceleration
= 2.7 * 3.25
= 8.775 N.
Using equations of motion,
distance,S = (13 * 4) - (1/2)(3.25)(4^2)
= 26 m
Workdone, W = force * distance
= 8.775 * 26
= 228.15 J
B.
Instantaneous power, P = Force *Velocity
= 8.775 * 13
= 114. 075 W
C.
t = 2 s,
Constant acceleration, a= (V-Vi)/t
= 13/2
= 6.5 m/s^2
Force = mass * acceleration
= 2.7 * 6.5
= 17.55 N
Instantaneous power, P = Force *Velocity
= 17.55 * 13
= 228.15 W.
= 114. 075 W.
A pressure gage connected to a tank reads 55 kPa at a location where the atmospheric pressure is 72.1 cmHg. The density of mercury is 13,600 kg/m3 . Calculate the absolute pressure in the tank.
Answer:
Explanation:
Given
Gauge Pressure of a tank is
[tex]P_{gauge}=55\ kPa[/tex]
at that place atmospheric Pressure is [tex]h=72.1\ cm\ of\ Hg[/tex]
Density of mercury [tex]\rho _{Hg}=13600\ kg/m^3[/tex]
Atmospheric Pressure in kPa is given by
[tex]P_{atm}=\rho _{Hg}\times g\times h[/tex]
[tex]P_{atm}=13600\times 9.8\times 0.721[/tex]
[tex]P_{atm}=96.09\ kPa[/tex]
and Absolute Pressure is summation of gauge pressure and atmospheric Pressure
[tex]P_{abs}=P_{gauge}+P_{atm}[/tex]
[tex]P_{abs}=55+96.09=151.09\ kPa[/tex]
How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?
Answer:
On moon time period will become 2.45 times of the time period on earth
Explanation:
Time period of simple pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex] ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth
As when we go to moon, acceleration due to gravity on moon is [tex]\frac{1}{6}[/tex] times os acceleration due to gravity on earth
So time period of pendulum on moon is equal to
[tex]T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}[/tex] --------eqn 2
Dividing eqn 2 by eqn 1
[tex]\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}[/tex]
[tex]T_{moon}=\sqrt{6}T=2.45T[/tex]
So on moon time period will become 2.45 times of the time period on earth
Final answer:
The period of a pendulum on the Moon would be longer because the Moon's gravity is weaker. To achieve the same one-second period as on Earth, a pendulum needs to be much shorter due to the Moon's 1/6th gravitational acceleration. Consequently, a pendulum's frequency would decrease if taken from Earth to the Moon.
Explanation:
The period of a simple pendulum is affected by the acceleration due to gravity, which is less on the Moon than on Earth. Hence, if you took a pendulum clock, like a grandfather clock, to the Moon, its pendulum would swing more slowly because of the Moon's weaker gravity. To maintain a steady tick-tock of one second per period on the Moon, the pendulum would need to be much shorter. A grandfather clock pendulum designed to have a two-second period on Earth with a length of 50 cm would need to be only 8.2 cm long on the Moon to achieve the same period, since the Moon's gravity is 1/6th that of Earth. Therefore, if a pendulum from Earth was taken to the Moon, its frequency would decrease because the acceleration due to gravity on the Moon is less than that on Earth.
If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the same?
Answer:
If the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
Explanation:
The total number of vibration or oscillation per unit time is called frequency of a waver. It is given by :
[tex]f=\dfrac{n}{t}[/tex]
n is the number of waves passing
t is the time taken
It is clear that the frequency is directly proportional to the number of waves. Hence, if the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
The solar system is of largely uniform composition. (T/F)
Answer:
False
Explanation:
Sun mass is dominating in Solar system as compared to other planets, asteroids and comets. Sun itself accounting for the 99.9% of the mass of the solar system. Hence the gravitational force exerted by the Sun dominates the other objects in the solar system. So we can conclude that solar system has non-uniform composition. The given statement is false
9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. The force of attraction / repulsion between the two charges will be: ______________________ 10) If one of the balloons has a mass of 0.084 kg, with what acceleration does it move toward or away from the other balloon? (calculate both magnitude AND direction) ________________________________________
Answer:
9. The force is a force of attraction and it is 2.95N
10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.
Explanation:
Parameters given:
Q1 = 3.4 * 10^-6C
Q2 = - 5.1 * 10^-6C
Distance between the two balloons = 23cm = 0.23m
9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:
F = kQ1Q2/r^2
F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2
F = (1.56 * 10^-1)/(5.29 * 10^-2)
F = - 2.95N
10. Assuming that Balloon A has a mass, m, of 0.084kg, then:
F = ma
Where a = acceleration
a = F/m
a = -2.95/0.084
a = - 35.12m/s^2
The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.
The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.
How does a person become "charged" as he or she shuffles across a carpet with bare feet on a dry winter day?
This process occurs because there is a contact between the carpet and the person's feet. Basically that contact generates the transfer of some electrons to the carpet on dry winter days.
In this way a person is charged when dragging bare feet on the carpet on a dry winter day.
Therefore, the net positive charge occurs on the surface of the carpet.
A person becomes charged when shuffling across a carpet due to the transfer of electrons from the feet to the carpet, leaving a net positive charge. The lack of humidity on a dry winter day allows the static charge to build up, leading to noticeable static shocks when touching a metal object. Humidity helps in dissipating the charge, making shocks less common on humid days.
When a person shuffles across a carpet with bare feet, they can become "charged" through a process known as charging by friction. This occurs when electrons are transferred from one surface to another due to the contact and relative motion between them. In this case, electrons move from the person's feet to the carpet, leaving the feet with a net positive charge.
Materials have different affinities for electrons, and when they come into close contact, the one with the higher affinity will take on electrons from the other. Since a dry winter day has low humidity, there is less moisture in the air to carry away the excess electrons. Therefore, the static charge you accumulate is less likely to be neutralized by the surrounding air, making static shocks more frequent and noticeable when you touch a metal object like a doorknob.
The reason for the shock is the rapid movement of electrons as they try to redistribute themselves to reach a state of electrical neutrality. When you touch a metal object, the excess electrons on your body rapidly transfer to the metal, causing the shock. On a humid day, the air's moisture helps electrons move away from your body more easily, preventing the build-up of a significant static charge.
You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of 5.00 m/s for another 120.0 m. For the total 240.0-m run, is your average velocity 4.00 m/s, greater than 4.00 m/s, or less than 4.00 m/s? Explain.
Answer:
Explanation:
Given
Speed while running towards east is [tex]v_1=3\ m/s[/tex]
Distance traveled in east direction [tex]x_1=120\ m[/tex]
For Another interval you run with velocity
[tex]v_2=5\ m/s[/tex]
[tex]x_2=240\ m[/tex]
Total displacement[tex]=x_1+x_2[/tex]
[tex]=120+120=240\ m[/tex]
Time for first interval
[tex]t_1=\frac{x_1}{v_1}=\frac{120}{3}[/tex]
[tex]t_1=\frac{120}{3}=40\ s[/tex]
Time for second interval
[tex]t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s[/tex]
total time [tex]t=t_1+t_2[/tex]
[tex]t=40+24=64\ s[/tex]
average velocity [tex]v_{avg}=\frac{x_1+x_2}{t}[/tex]
[tex]v_{avg}=\frac{240}{64}=3.75\ m/s[/tex]
Therefore average velocity is less than [tex]4 m/s[/tex]
An excited hydrogen atom emits light with a wavelength of 486.4 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin?
Answer:
The electron began in the quantum level of 4
Explanation:
Using the formula of wave number:
Wave Number = 1/λ = Rh(1/n1² - 1/n2²)
where,
Rh = Rhydberg's Constant = 1.09677 x 10^7 /m
λ = wavelength of light emitted = 486.4 nm = 486.4 x 10^-9 m
n1 = final shell = 2
n2 = initial shell = ?
Therefore,
1/486.4 x 10^-9 m = (1.09677 x 10^7 /m)(1/2² - 1/n2²)
1/4 - 1/(486.4 x 10^-9 m)(1.09677 x 10^7 /m) = 1/n2²
1/n2² = 0.06082
n2² = 16.44
n2 = 4
The principal quantum level in which the electron began is 5.
Given the following data:
Final transition = 2Wavelength = 486.4 nm = [tex]486.4 \times 10^{-9}\;m[/tex]Rydberg constant = [tex]1.09 \times 10^7\; m^{-1}[/tex]
To determine the principal quantum level in which the electron began, we would use the Rydberg equation:
Mathematically, the Rydberg equation is given by the formula:
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} -\frac{1}{n_i^2})[/tex]
Where:
[tex]\lambda[/tex] is the wavelength.R is the Rydberg constant.[tex]n_f[/tex] is the final transition.[tex]n_i[/tex] is the initial transition.Substituting the parameters into the formula, we have;
[tex]\frac{1}{486.4 \times 10^{-9}} = 1.09 \times 10^7(\frac{1}{2^2} -\frac{1}{n_i^2})\\\\\frac{1}{486.4 \times 10^{-9} \times 1.09 \times 10^7}=\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{5.3018} =\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{n_i^2}=\frac{1}{4}-\frac{1}{5.3018}\\\\\frac{1}{n_i^2}=0.25-0.1886\\\\\frac{1}{n_i^2}=0.0614\\\\n_i^2=\frac{1}{0.0614} \\\\n_i=\sqrt{16.29} \\\\n_i=4.0[/tex]
Initial transition = 4.0
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Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
Answer:
Explanation:
Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?
Answer:
q = 8.85 x 10⁻¹¹ C
Explanation:
given,
Electric field, E = 1.18 N/C
distance, r = 0.822 m
Charge magnitude = ?
using formula of electric field.
[tex]E = \dfrac{kq}{r^2}[/tex]
k is the coulomb constant
[tex]q= \dfrac{Er^2}{k}[/tex]
[tex]q= \dfrac{1.18\times 0.822^2}{9\times 10^9}[/tex]
q = 8.85 x 10⁻¹¹ C
The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C
In reaching her destination, a backpacker walks with an average velocity of 1.20 m/s, due west. This average velocity results, because she hikes for 5.63 km with an average velocity of 2.33 m/s due west, turns around, and hikes with an average velocity of 0.374 m/s due east. How far east did she walk (in kilometers)?
Answer:
[tex] x_2 = 648.46\ m[/tex]
Explanation:
given,
Average velocity due west = 1.20 m/s
case 1
Distance moved in west, x₁ = 5.63 km
speed due west, v₁ = 2.33 m/s
Case 2
Distance moved in east = x₂
speed due east, v₂ = 0.374 m/s
total distance = x₁ + x₂ = 5.62
total time = t₁ + t₂ = 5630/2.33 + x₂/0.374
now,
[tex]average\ velocity = \dfrac{total\ distance}{total\ time}[/tex]
[tex]-1.20= \dfrac{-5630 + x_2}{\dfrac{5630}{2.33}+\dfrac{x_2}{0.374}}[/tex]
negative sign is used because we want the distance in east but velocity is in west.
[tex] - 2900 - 3.21 x_2 = -5630 + x_2[/tex]
[tex]-4.21 x_2 = -2730[/tex]
[tex] x_2 = 648.46\ m[/tex]
The distance she walked in east is equal to 648.46 m
The plates of a parallel-plate capacitor are 3.50 mm apart, and each carries a charge of magnitude 75.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×10^6 V/m.a. What is the potential difference between the plates?b. What is the area of each plate?c. What is the capacitance?
Answer:
Vab =17.5kV
A = 16.9 cm2
C = 4.27pF
Explanation:
a) Find the voltage difference:
Vab = Ed
E Electric field
d distance between plates
Vab potential difference
d = 3.5mm
= 3.5 * 10^(-3) m
Q = 75.0nC
= 75 * 10^(-9)
E = 5.00 * 10^6 V/m
Vab = (5.00 * 10^6) * (3.5 * 10^(-3))
= 17.5 * 10^3 V
=17.5kV
b. What is the area of the plate?
The relation between the electric field and area is given as:
E = Q/(ϵ0 * A)
A = Q/(ϵ0 *E)
Where ϵ0 is the electric constant and equals 8.854 × 10^ (-12) C2/N•m2
A = 75 * 10^ (-9) / (8.854 × 10^ (-12) (5.00 * 10^6)
= 1.69 X 10^ (-3) m2
= 16.9 cm2
c. Find the capacitance
The equation relating capacitance, area of plate and plate distance is given by:
C = ϵ0 A/d
plug in the values of d, ϵ0 and A above to get the capacitance:
C = (8.854 × 10^ (-12) * 1.69 X 10^ (-3) / 3.5 * 10^ (-3)
= 4.27 * 10^ (-12) F
= 4.27pF
True or False? The superposition or overlapping of two waves always results in destructive interference between the different waves.
Answer:
Explanation:
False
When two waves overlap or superimpose over each other then they either undergo Constructive or destructive interference.
waves are the disturbance created by a force and add up to gives constructive interference when they are in the same line i.e. in the same phase.
When these disturbances are in the opposite phase then they superimpose to give destructive interference where the amplitude of the resulting wave will be much smaller as compared to original waves.
Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 92.2mV. The resistivity of the membrane material is 1.30 x 10^7 ohms*m
PLEASE SHOW WORK!
a) Determine the amount of current that flows through this portion of the membrane
Answer: _____A
b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do no change
increase by factor of 2
decrease by factor of 8
decrease by factor of 2
decrease by a factor of 4
increase by factor of 4
The amount of current that flows through this given portion of a cell membrane, calculated using Ohm's law and the properties of the membrane, is 1.60 µA. If the side dimensions of the membrane are halved, the current will decrease by a factor of 4.
Explanation:The relevant concept needed to answer these questions is Ohm's Law, defined as Voltage = Current x Resistance. In this context, Resistance = Resistivity x (Thickness/Area) and the area is a square.
a) Determine the amount of current that flows through this portion of the membrane:
First, calculate the resistance: R = ρ x (Thickness/ Area)
Remove the micrometers units of the area and convert it into meters to match the ρ units. So, you get an area of 1.3 x 10^-6 m x 1.3 x 10^-6 m = 1.69 x 10^-12 m^2. Then, R = 1.30 x 10^7 Ω*m x (7.50 x 10^-9 m / 1.69 x 10^-12 m^2) = 57.404 Ω.
By plugging the calculated resistance and given voltage into Ohm's Law, we can find the current: I = V/R = 92.2 x 10^-3 V / 57.4 Ω = 1.60 μA
b) By what factor does the current change if the side dimensions of the membrane portion is halved:If the side dimensions are halved, the area of the membrane becomes one-fourth of the original, thus the resistance increases by a factor of 4. According to Ohm's Law, as resistance increases, the current decreases, meaning that if the resistance is multiplied by 4, the current will decrease by a factor of 4.
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a) Therefore, the amount of current that flows through this portion of the membrane is approximately [tex]\({1.60 \times 10^{-6} \, \text{A}} \)[/tex]. b) The correct answer is decrease by a factor of 5208. The current decreases by a factor of approximately 5208.
Part (a): Determine the amount of current that flows through this portion of the membrane
To find the current ( I ) flowing through the membrane portion, we use Ohm's law and the given potential difference ( V ) across the membrane.
1. Calculate the resistance ( R ) of the membrane:
The resistivity [tex](\( \rho \))[/tex] is given as [tex]\( 1.30 \times 10^7 \) ohms\·m.[/tex]
First, calculate the cross-sectional area ( A ) of the membrane portion:
[tex]\[ A = 1.3 \, \mu \text{m} \times 1.3 \, \mu \text{m} = (1.3 \times 10^{-6} \, \text{m})^2 = 1.69 \times 10^{-12} \, \text{m}^2 \][/tex]
Then, calculate the resistance ( R ):
[tex]\[ R = \frac{\rho \cdot L}{A} \][/tex]
[tex]\[ R = \frac{1.30 \times 10^7 \, \text{ohm} \cdot \text{m} \cdot 7.50 \times 10^{-9} \, \text{m}}{1.69 \times 10^{-12} \, \text{m}^2} \][/tex]
[tex]\[ R = \frac{9.75 \times 10^{-2}}{1.69 \times 10^{-12}} \approx 5.77 \times 10^7 \, \text{ohms} \][/tex]
2. Calculate the current ( I ):
Ohm's law states [tex]\( I = \frac{V}{R} \).[/tex]
Given potential difference [tex]\( V = 92.2 \, \text{mV} = 92.2 \times 10^{-3} \, \text{V} \):[/tex]
[tex]\[ I = \frac{92.2 \times 10^{-3} \, \text{V}}{5.77 \times 10^7 \, \text{ohms}} \approx 1.60 \times 10^{-6} \, \text{A} \][/tex]
Part (b): By what factor does the current change if the side dimensions of the membrane portion is halved?
If the side dimensions of the membrane portion are halved, the cross-sectional area ( A ) of the membrane will decrease by a factor of ( 4 ) (since both length and width are halved).
1. New cross-sectional area ( A' ):
[tex]\[ A' = \left( \frac{1.3 \, \mu \text{m}}{2} \right) \times \left( \frac{1.3 \, \mu \text{m}}{2} \right) = \left( \frac{1.3}{2} \times 10^{-6} \, \text{m} \right)^2 = 0.325 \times 10^{-12} \, \text{m}^2 \][/tex]
2. New resistance ( R' ):
Using the same resistivity [tex]\( \rho \)[/tex] and thickness ( L ):
[tex]\[ R' = \frac{\rho \cdot L}{A'} = \frac{1.30 \times 10^7 \cdot 7.50 \times 10^{-9}}{0.325 \times 10^{-12}} \approx 3.00 \times 10^8 \, \text{ohms} \][/tex]
3. New current ( I' ):
[tex]\[ I' = \frac{V}{R'} = \frac{92.2 \times 10^{-3}}{3.00 \times 10^8} \approx 3.07 \times 10^{-10} \, \text{A} \][/tex]
4. Calculate the factor by which the current changes:
[tex]\[ \frac{I'}{I} = \frac{3.07 \times 10^{-10}}{1.60 \times 10^{-6}} \approx 1.92 \times 10^{-4} \][/tex]
Since the current decreases, we consider the reciprocal:
[tex]\[ \frac{I}{I'} \approx \frac{1}{1.92 \times 10^{-4}} \approx 5208 \][/tex]
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. The properties of steam at 220°C are given as follows: vf = 0.001190 m3/kg and vg = 0.08609 m3/kg.
Answer:
a) P = 2319.6[kPa]; b) 2.6%
Explanation:
Since the problem data is not complete, the following information is entered:
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated mixture.
From the information provided in the problem we can say that you have a mixture of liquid and steam.
a) Using the steam tables we can see (attached image) that the saturation pressure at 220 °C is equal to:
[tex]P_{sat} =2319.6[kPa][/tex]
[tex]v_{f}=0.001190[m^{3}/hr]\\v_{g}=0.08609[m^{3}/hr]\\[/tex]
b) Since the specific volume of the gas and liquid is known, we can find the mass of each phase using the following equation:
[tex]m_{f}=\frac{V_{f} }{v_{f} } \\m_{g}=\frac{V_{g} }{v_{g} } \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\[/tex]
We know that the volume of the fluid is equal to:
[tex]V_{f}=1/3*V_{total} \\V_{total}=1.78[m^3]\\[/tex]
Now we can find the mass of the gas and the liquid.
[tex]m_{f}=\frac{1/3*1.78}{0.001190} \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg][/tex]
The total mass is the sum of both
[tex]m_{total} =m_{g} + m_{fluid} \\m_{total} = 498.6 + 13.78\\m_{total} = 512.38[kg][/tex]
The quality will be equal to:
[tex]x = \frac{m_{g} }{m_{T} }\\ x= \frac{13.78}{512.38} \\x = 0.026 = 2.6%[/tex]
Is the electric-field magnitude between the plates larger or smaller than that for the original capacitor?
Answer:
The magnitude of the electric field will decrease
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d
C1=K(ϵ0A/d)
C1=KC
Where C is the capacitance with no dielectric.
The new capacitance is k times the capacitance of the capacitor without dielectric slab.
This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=KC. Inserting this into equation 2
CV = KCV1
V1 = (CV)/KC
V/K
This implies the voltage decreases when a dielectric is used within the plate.
The relationship between electric field and potential voltage is a linear one
V= Ed
Therefore the electric field will decrease
Final answer:
The capacitance of a parallel-plate capacitor with non-uniform electric field due to increased plate separation will be less than the ideal scenario calculated by C = € A/d.
Explanation:
When the separation of the plates in a parallel-plate capacitor is not small enough to maintain a uniform electric field, the field lines will bulge outwards at the edges. This bulging implies that some of the field lines that emanate from one plate do not reach the opposite plate, reducing the effective area over which the field is acting. Consequently, this non-uniform field means that the capacitance of the capacitor will be less than the ideal calculation using the formula C = € A/d, where C is the capacitance, € is the permittivity of the medium between the plates, A is the plate area, and d is the separation between the plates.
A block of mass M M is placed on a semicircular track and released from rest at point P P , which is at vertical height H 1 H1 above the track’s lowest point. The surfaces of the track and block are considered to be rough such that a coefficient of friction exists between the track and the block. The block slides to a vertical height H 2 H2 on the other side of the track. How does H 2 H2 compare to H 1 H1 ?
Answer:
Explanation:
A block of mass M is placed on a semicircular track and released from rest at point P , which is at vertical height H₁ above the track’s lowest point.
Its initial potential energy = mgH₁
Kinetic energy = 0
Total energy = mgH₁
When block slides to a vertical height H₂ on the other side of the track
Its final potential energy = mgH₂
Kinetic energy = 0
Total final energy = mgH₂
As negative work is done by frictional force while block moves ,
final energy < initial energy
mgH₂ < mgH₁
H₂ < H₁
H₂ will be less than H₁ .
What is the strength of an electric field that will balance the weight of a 9.6 g plastic sphere that has been charged to -9.2 nC ? Express your answer to two significant figures and include the appropriate units.
The strength of an electric field that will balance the weight is 1.023 × 10⁷ N/C.
What is electric field?An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. Additionally, it refers to a system of charged particles' physical field.
Electric charges and time-varying electric currents are the building blocks of electric fields.
The strength of an electric field that will balance the plastic sphere is = weight of the object/charge on the object
= ( 9.6 ×10⁻³×9.8)/(9.2×10⁻⁹) N/C
= 1.023 × 10⁷ N/C
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A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is required to pump the water out of the pool when it is full? Use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity.
Final answer:
The problem requires calculating the work done to pump water out of a full swimming pool using given dimensions, the density of water, and gravity.
Explanation:
The question involves finding the amount of work required to pump the water out of a swimming pool when it is full. The dimensions of the pool are given, along with the density of water and the acceleration due to gravity. Using the density of water (1000 kg/m3), the volume of the pool can be calculated to determine the total mass of the water. The work done in pumping the water is found by multiplying the mass by the gravitational constant (9.8 m/s2) and the vertical distance the water needs to be moved (2.2 m, which is the uniform depth of the pool). This distance can be different depending on the location of the pump, but for this problem, we assume the water is being pumped from the very bottom.
What is the relationship between wavelength, wave frequency, and wave velocity?
Relation Between Velocity And Wavelength
Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance travelled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.
Velocity And Wavelength
For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-
V=f×λ
Where,
V is the velocity of the wave measure using m/s.
f is the frequency of the wave measured using Hz.
λ is the wavelength of the wave measured using m.
Velocity and Wavelength Relationtion
Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.
Given by:
V∝λ
Example:
For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.
For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.
Hope you understood the relation between wavelength and velocity of a wave. You may also want to check out these topics given below!
Relation between phase difference and path difference
Relation Between Frequency And Velocity
Relation Between Escape Velocity And Orbital Velocity
Relation Between Group Velocity And Phase Velocity
The relationship between wavelength, wave frequency, and wave velocity is described by the equation v = fλ. Wavelength and frequency are inversely proportional given a constant wave velocity – high frequency correlates with short wavelength and vice versa.
Explanation:In Physics, there's a mathematical relationship between wavelength, wave frequency, and wave velocity for any type of wave motion. This relationship is often stated as v = fλ, where v is wave velocity, f is the frequency of the wave, and λ is the wavelength. The wavelength is the distance between identical parts of the wave, while the velocity is the speed at which the disturbance moves, and the frequency is the rate of oscillation of the wave.
When you look at this formula, it becomes clear that if the wave velocity (v) is constant, a wave with a longer wavelength (λ) will have lower frequency (f). On the other side, higher frequency means shorter wavelength. This is because frequency and wavelength are inversely proportional in the given formula.
Example
For instance, the speed of light in vacuum is a constant value (approximately 3.00×108 m/s). So, if a certain light wave has a larger wavelength, its frequency will be lower to ensure this speed remains consistent.
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A square is 1.0 m on a side. Point charges of +4.0 µC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 µC and -3.0 µC. What is the potential (relative to infinity) at the midpoint of the square?
Answer:
[tex]V = 1.44\times 10^{5}~V[/tex]
Explanation:
The electric potential can be found by using the following formula
[tex]V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
Applying this formula to each charge gives the total potential.
[tex]V = V_1 + V_2 + V_3 + V_4\\V = \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2} - \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2}\\V = \frac{16\times 10^{-6}}{4\pi\epsilon_0}\\V = 1.44\times 10^{5}~V[/tex]
Since the potential is a scalar quantity, it is safe to sum all the potentials straightforward. And since they all placed on the corners of a square, +3 and -3 μC charges cancel out each other.
An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is the diameter of the filament reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.
[tex]R\propto\dfrac{1}{D}[/tex]
The power is inversely proportional to the resistance.
[tex]P\propto\dfrac{1}{R}[/tex]
[tex]P\propto D^2[/tex]
We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter
[tex]\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}[/tex]
Put the value into the formula
[tex]\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}[/tex]
[tex]\dfrac{D_{i}}{D_{f}}=0.95[/tex]
[tex]D_{i}=0.95 D_{f}[/tex]
Hence, The factor of the diameter is 0.95.
Answer:
Explanation:
Po = 60 W
P = 54.3 W
Let the initial diameter of the filament is do and the final diameter of the filament is d.
Let the voltage is V and the initial resistance is Ro and the final resistance is R.
The formula for power is given by
P = V²/R
The resistance of the filament is inversely proportional to the square of the diameter of the filament. As voltage is constant so the power is
Power α diameter²
So, initial power is
Po α do² ..... (1)
Final power is
P α d² ..... (2)
Divide equation (2) by equation (1), we get
P / Po = d² / do²
54.3 / 60 = d² / do²
d² / do² = 0.905
d = 0.95 d
Thus, the diameter of the filament is reduced to a factor of 0.95 .
The acceleration of a bus is given by ax(t) = αt, where α = 1.2 m/s3. (a) If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its velocity at time t = 2.0 s? (b) If the bus’s position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s? (c) Sketch ay-t, vy-t , and x-t graphs for the motion.
Answer:
(a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
Explanation:
Given that,
[tex]\alha=1.2\ m/s^3[/tex]
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
[tex]a_{x(t)}=\alpha t[/tex]
We need to calculate the velocity
Using formula of acceleration
[tex]a=\dfrac{dv}{dt}[/tex]
On integrating
[tex]\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}[/tex]
Put the value into the formula
[tex]v-v_{0}=1.2\int_{0}^{t}{t dt}[/tex]
[tex]v-v_{0}=0.6t^2[/tex]
[tex]v=v_{0}+0.6t^2[/tex]
Put the value into the formula
[tex]v_{0}=5.0-0.6\times(1.0)^2[/tex]
[tex]v_{0}=4.4\ m/s[/tex]
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
[tex]v=4.4+0.6\times(2.0)^2[/tex]
[tex]v=6.8\ m/s[/tex]
(b). If the bus’s position at time t = 1.0 s is 6.0 m,
We need to calculate the position
Using formula of velocity
[tex]v=\dfrac{dx}{dt}[/tex]
On integrating
[tex]\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}[/tex]
[tex]x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}[/tex]
[tex]x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3[/tex]
[tex]x=1.4\ m[/tex]
The position at t = 2.0 s
[tex]x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3[/tex]
[tex]x=11.8\ m[/tex]
Hence, (a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs