Final answer:
The question is about using the single slit diffraction formula to calculate the angles at which the third dark fringe appears for two different wavelengths and the width of the central bright fringe for each wavelength, involving physics and high school level understanding.
Explanation:
The student's question relates to the concept of single slit diffraction, which is a phenomenon in physics where light spreads out after passing through a narrow opening, resulting in a characteristic pattern. Specifically, the angles at which dark fringes appear and the width of the central bright fringe are calculated for two different wavelengths using the formulas for single slit diffraction.
For part (a), to calculate the angles for the third dark fringe for each wavelength, we can use the formula dsin(θ) = mλ, where d is the slit width, θ is the angle of the dark fringe, m is the order of the dark fringe, and λ is the wavelength of the light. The third dark fringe corresponds to m = 3 for this calculation. For part (b), the width of the central bright fringe is calculated by considering the angles to the first dark fringes on either side of the central peak and then using trigonometry to determine the width on the screen.
Suppose a photon with an energy of 1.60 eV strikes a piece of metal. If the electron that it hits loses 0.800 eV leaving the metal, what is the kinetic energy of the electron once it has broken free of the metal surface?
To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.
From the theory we could consider that the energy change is subject to
[tex]\Delta E = E_0 -W_f[/tex]
Where
[tex]E_0 =[/tex]Initial Energy
[tex]W_f =[/tex] Energy loses
Replacing we have that
[tex]\Delta E = 1.6-0.8[/tex]
[tex]\Delta E = 0.8eV[/tex]
Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV
A glass plate (n = 1.60) is covered with a thin, uniform layer of oil (n = 1.29). A light beam of variable wavelength from air is incident normally on the oil surface. Observation of the reflected beam shows constructive interference at 511 nm. Determine the minimum non-zero thickness of the oil film.
Answer:
The thickness of the oil film is 198 nm.
Explanation:
Given that,
Refractive index of glass plate = 1.60
Refractive index of oil = 1.29
Wavelength = 511 nm
We need to calculate the thickness of the oil film
Using formula of path difference
[tex]2nt=k\lambda[/tex]
[tex]t=\dfrac{k\times\lambda}{2n}[/tex]
Where, n = refractive index
t = thickness
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]t=\dfrac{1\times511\times10^{-9}}{2\times1.29}[/tex]
[tex]t=198\times10^{-9}\ m[/tex]
[tex]t= 198\ nm[/tex]
Hence, The thickness of the oil film is 198 nm.
A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m.
While at this equilibrium position, the mass is then given an initial push downward at v = 4.1 m/s. The block oscillates on the spring without friction.
1)What is the spring constant of the spring?
2)What is the oscillation frequency?
3)After t = 0.42 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?
5)At t = 0.42 s what is the magnitude of the net force on the block?
1. Spring constant k is 233.17 N/m.
2. Oscillation frequency f is 0.924 Hz.
3. Speed of the block after 0.42 s is 2.62 m/s.
4. Magnitude of the maximum acceleration is 23.8 m/s².
5. Magnitude of the net force at 0.42 s is 105.02 N.
[tex]\\\( \omega_i \)[/tex]is the initial angular velocity (given as 90.0 rpm), Let's solve the problem step-by-step.
1. What is the spring constant of the spring?
The spring constant k can be determined using Hooke's Law F = kx , where F is the force exerted by the spring and x is the displacement.
At equilibrium, the force exerted by the spring equals the weight of the block:
F = mg
Given:
m = 6.9 kg
g = 9.8 m/s²
x = 0.29 m
We calculate the force:
F = 6.9 kg * 9.8 m/s² = 67.62 N
Now, using F = kx :
k = F/x = 67.62 N/0.29 m = 233.17 N/m
Spring constant k is 233.17 N/m
2. What is the oscillation frequency?
The frequency [tex]\( f \)[/tex] can be found using the formula for the angular frequency [tex]\( \omega \):[/tex]
[tex]\[ \omega = \sqrt{\frac{k}{m}} \][/tex]
Given:
[tex]\[ k = 233.17 \text{ N/m} \]\[ m = 6.9 \text{ kg} \][/tex]
Calculate [tex]\( \omega \):[/tex]
[tex]\[ \omega = \sqrt{\frac{233.17 \text{ N/m}}{6.9 \text{ kg}}} = \sqrt{33.8 \text{ s⁻²}} = 5.81 \text{ rad/s} \][/tex]
The frequency[tex]\( f \)[/tex]is:
[tex]\[ f = \frac{\omega}{2\pi} = \frac{5.81 \text{ rad/s}}{2\pi} \approx 0.924 \text{ Hz} \][/tex]
Oscillation frequency f is 0.924 Hz
3. After t = 0.42 s, what is the speed of the block?
The speed of the block in simple harmonic motion can be described by:
[tex]\[ v(t) = \omega A \cos(\omega t + \phi) \][/tex]
The amplitude A and phase [tex]\( \phi \)[/tex]are determined from initial conditions.
The initial conditions:
- Initial position [tex]\( x(0) = 0 \)[/tex](at equilibrium)
- Initial velocity [tex]\( v(0) = -4.1 \text{ m/s} \)[/tex]
Since [tex]\( x(0) = A \cos(\phi) = 0 \), \(\phi = \pi/2\) (or 90°).[/tex]
Then, [tex]\( v(0) = -\omega A \sin(\phi) = -\omega A \):[/tex]
[tex]\[ -4.1 = -5.81 A \]\[ A = \frac{4.1}{5.81} \approx 0.705 \text{ m} \][/tex]
Now, calculate [tex]\( v(t) \) at \( t = 0.42 \text{ s} \):[/tex]
[tex]\[ v(0.42) = 5.81 \times 0.705 \times \cos(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ v(0.42) = 5.81 \times 0.705 \times \cos(2.44 + 1.57) \]\[ v(0.42) = 4.096 \times \cos(4.01) \]\[ v(0.42) = 4.096 \times -0.64 = -2.62 \text{ m/s} \][/tex]
The speed of the block after 0.42 s is 2.62 m/s
4. What is the magnitude of the maximum acceleration of the block?
The maximum acceleration [tex]\( a_{max} \)[/tex] occurs at the maximum displacement [tex]\( A \):[/tex]
[tex]a_max} = \omega[/tex]² A
Given:
[tex]\[ \omega = 5.81 \text{ rad/s} \]\[ A = 0.705 \text{ m} \][/tex]
Calculate:
[tex]\[ a_{max} = (5.81)^2 \times 0.705 = 33.76 \times 0.705 = 23.8 \text{ m/s^2} \][/tex]
The magnitude of the maximum acceleration is 23.8 m/s²
5. At t = 0.42 s, what is the magnitude of the net force on the block?
The net force ( F(t) ) at any time ( t ) can be found using:
[tex]\[ F(t) = ma(t) \][/tex]
Where [tex]\( a(t) = -\omega^2 x(t) \). From \( x(t) = A \sin(\omega t + \phi) \):[/tex]
[tex]\[ x(0.42) = 0.705 \sin(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ x(0.42) = 0.705 \sin(2.44 + 1.57) \]\[ x(0.42) = 0.705 \sin(4.01) \]\[ x(0.42) = 0.705 \times -0.64 = -0.451 \text{ m} \][/tex]
Now, [tex]\( a(0.42) = -\omega^2 x(0.42) \):[/tex]
[tex]\[ a(0.42) = - (5.81)^2 \times (-0.451) \]\[ a(0.42) = -33.76 \times -0.451 = 15.22 { m/s^2}[/tex]
Then, the net force:
[tex]\[ F(0.42) = 6.9 \text{ kg} \times 15.22 \text{ m/s^2} = 105.02 \text{ N} \][/tex]
The magnitude of the net force at 0.42 s is 105.02 N
A wood salvage company is hoisting an old tree trunk off the bottom and out of a lake. The cable from the hoist is tied around the log above its center of mass. The hoist applies a force of 9,800 N to the cable to suspend the log in the lake water (FT water), and a force of 29,000 N to suspend the log above the lake surface (FT air). What are the volume and density of the log? Assume the lake water has a density of 1,007 kg/m3.
To solve this problem it is necessary to resort to the concepts expressed in the Buoyancy Force.
The buoyancy force is given by the equation
[tex]F = \rho Vg[/tex]
Where,
[tex]\rho =[/tex] Density
V =Volume
g = Gravitational Acceleration
PART A) From the given data we can find the volume, so
[tex]9800 = 1007*V*9.8[/tex]
[tex]V = 0.99m^3[/tex]
PART B) The mass can be expressed from the Newton equation in which
[tex]F = mg[/tex]
Where,
m = mass
g = Gravitational acceleration
Replacing with our values we have that
[tex]29000 = m*9.8[/tex]
[tex]m = 2959.18Kg[/tex]
Therefore the Density can be calculated with the ratio between the Volume and Mass
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{2959.18Kg}{0.99m^3}[/tex]
[tex]\rho = 2989.074kg/m^3[/tex]
Therefore the Density of the log is [tex]2989.074kg/m^3[/tex]
A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1.25 m radius of the balloon. The balloon is filled with helium (He) at a temperature of 345 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m³ and the molar mass of helium is 4.0026×10-3 kg/mol. Find the absolute pressure of the helium gas.
Answer:
563712.04903 Pa
Explanation:
m = Mass of material = 3.3 kg
r = Radius of sphere = 1.25 m
v = Volume of balloon = [tex]\frac{4}{3}\pi r^3[/tex]
M = Molar mass of helium = [tex]4.0026\times 10^{-3}\ kg/mol[/tex]
[tex]\rho[/tex] = Density of surrounding air = [tex]1.19\ kg/m^3[/tex]
R = Gas constant = 8.314 J/mol K
T = Temperature = 345 K
Weight of balloon + Weight of helium = Weight of air displaced
[tex]mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg[/tex]
Mass of helium is 6.4356 kg
Moles of helium
[tex]n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489[/tex]
Ideal gas law
[tex]P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa[/tex]
The absolute pressure of the Helium gas is 563712.04903 Pa
Final answer:
To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere. Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure.
Explanation:
To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, we need to solve for P. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere and use the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius. Given that the radius is 1.25 m and the volume is known, we can calculate the number of moles of helium using the ideal gas law and the molar mass of helium.
Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure. Remember to convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature.
A particle with charge q and kinetic energy KE travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for its speed v and its mass m. (Use any variable or symbol stated above as necessary.) (a) expression for its speed (Do not use m in your answer.) v = (b) expression for its mass (Do not use v in your answer.) m =
Answer:
a)[tex]v=\dfrac{2.KE}{qBR}[/tex]
b)[tex]m=\dfrac{(qBR)^2}{2.KE}[/tex]
Explanation:
Given that
Charge = q
Magnetic filed = B
Radius = R
We know that kinetic energy KE
[tex]KE=\dfrac{1}{2}mv^2[/tex] ----------1
m v² = 2 .KE
The magnetic force F = q v B
Radial force
[tex]Fr=\dfrac{1}{R}mv^2[/tex]
For uniform force these two forces should be equal
[tex]q v B=\dfrac{1}{r}mv^2[/tex]
q v B R =m v²
q v B R = 2 .KE
[tex]v=\dfrac{2.KE}{qBR}[/tex]
Now put the velocity v in the equation
[tex]KE=\dfrac{1}{2}mv^2[/tex]
[tex]m=\dfrac{2 .KE}{v^2}[/tex]
[tex]m=\dfrac{2.KE}{\left(\dfrac{2.KE}{qBR}\right)^2}[/tex]
[tex]m=\dfrac{(qBR)^2}{2.KE}[/tex]
Final answer:
To find the speed and mass of a charged particle moving in a magnetic field, we use its uniform circular motion to derive formulas for speed (v = qBR/m) and mass (m = q^2B^2R^2/(2KE)). These expressions help us understand the relationship between the particle's physical properties and its motion within the magnetic field.
Explanation:
To find the expressions for the speed v and mass m of a charged particle moving in a circular path of radius R inside a uniform magnetic field, we utilize the motion equations for a particle undergoing uniform circular motion due to a magnetic force. The magnetic force provides the centripetal force required for this motion and is given by qvB, where q is the charge, v is the velocity, and B is the magnetic field strength. This force is equal to the centripetal force needed for circular motion, which is given by mv2/R.
(a) Rearranging the formula qvB = mv2/R to solve for v, we get:
v = qBR/m
(b) To find the mass m, we use the particle’s kinetic energy KE and the velocity from part (a). Since KE = 1/2 mv2 and the value of v is qBR/m, we can substitute it to get:
m = q2B2R2/(2KE)
The units of these expressions can also be derived to ensure they're correct. Additionally, if we consider changes to the magnetic field's strength or the particle's charge, we can infer how these changes affect the radius R using the derived formulas.
The molecular clouds circling the active nucleus of the galaxy M106 orbit at a speed of about 1000 km/s, with an orbital radius of 0.49 light-year = 4.8×1015 meters. Part A Use the orbital velocity law Mr=r×v2G to calculate the mass of the central black hole. Give your answer in kilograms.
Answer:
The mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]
Explanation:
The Universal law of gravitation shows the interaction of gravity between two bodies:
[tex]F = G\frac{Mm}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.
For this particular case M is the mass of the central black hole and m is the mass of the molecular cloud. Since it is a circular motion the centripetal acceleration will be:
[tex]a = \frac{v^{2}}{r}[/tex] (2)
Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):
[tex]ma = G\frac{Mm}{r^{2}}[/tex]
By replacing (2) in equation (1) it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)
Therefore, the mass of the central black hole can be determined if M is isolated from equation (3):
[tex]M = \frac{rv^{2}}{G}[/tex] (4)
Equation 4 it is known as the orbital velocity law.
Where M is the mass of the central black hole, r is the orbital radius, v is the orbital speed and G is the gravitational constant.
Before replacing the orbital speed in equation 4 it is necessary to make the conversion from kilometers to meters:
[tex]1000 \frac{km}{s} x \frac{1000 m}{1 km}[/tex] ⇒ [tex]1000000 m/s[/tex]
Then, equation 4 can be finally used:
[tex]M = \frac{(4.8x10^{15} m)(1000000 m/s)^{2}}{(6.67x10^{-11} N.m^{2}/Kg^{2})}[/tex]
[tex]M = 7.19x10^{37} Kg[/tex]
Hence, the mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]
To calculate the mass of the central black hole, we can use the orbital velocity law, Mr = r×v^2/G. Given the orbital radius and velocity of the molecular clouds, we can plug these values into the equation to find the mass of the black hole.
Explanation:To calculate the mass of the central black hole, we can use the orbital velocity law: Mr=r×v^2/G. In this equation, Mr represents the mass of the black hole, r is the orbital radius, v is the orbital velocity, and G is the gravitational constant. Given that the orbital radius is 0.49 light-years (4.8x10^15 meters) and the orbital velocity is 1000 km/s, we can plug these values into the equation to find the mass of the black hole.
First, let's convert the orbital velocity from km/s to m/s: 1000 km/s = 1000 x 1000 m/s = 1,000,000 m/s. Now, we plug in the values to the equation:
Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11 m^3⋅kg^−1⋅s^−2)
Simplifying the expression, we get:
Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11) kg
After calculating, the mass of the central black hole is approximately 3.23x10^37 kg.
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A 15.0-Ω resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed switch.
(a) At 2.00 ms after the switch is opened the current has decayed to 0.210 A. Calculate the inductance of the coil.
(b) Calculate the time constant of the circuit.
(c) How long after the switch is opened will the current reach 1.00% of its original value?
The inductance of the coil is 30 mH. The time constant of the circuit is 2.00 ms. It will take approximately -4.60 ms for the current to reach 1.00% of its original value.
Explanation:To find the inductance of the coil, we can use the formula for the time constant in an RL circuit, T = L/R, where T is the time constant, L is the inductance, and R is the resistance. In this case, the resistance is given as 15.0 Ω, and the time it takes for the current to decay to 0.210 A is given as 2.00 ms. Plugging in these values, we can solve for L:
T = L/R
L = T × R
L = (2.00 ms) × (15.0 Ω)
L = 30 mH
Therefore, the inductance of the coil is 30 mH.
The time constant of the circuit is given by the formula T = L/R. Plugging in the values of L and R from part (a), we can solve for T:
T = L/R
T = (30 mH) / (15.0 Ω)
T = 2.00 ms
Therefore, the time constant of the circuit is 2.00 ms.
To find the time it takes for the current to reach 1.00% of its original value, we can use the formula for the time constant in an RL circuit, T = L/R. In this case, the resistance is given as 15.0 Ω, and we want to find the time when the current is 1.00% of its original value. Let's call this time T₁. Plugging in these values, we can solve for T₁:
T = L/R
T₁ = T × (ln(I1/I0))
T₁ = (2.00 ms) × (ln(0.01/1.00))
T₁ = -4.60 ms
Therefore, it will take approximately -4.60 ms for the current to reach 1.00% of its original value.
A heater 0.015 m in diameter and 0.3 m long is submerged horizontally in oil at 20 oC. To avoid oil fumes, the heater surface should not exceed Ts= 150 oC. Calculate the maximum power that should be supplied to the heater.
Answer:
The maximum power is 23.89 k watt.
Explanation:
Given that,
Diameter = 0.015 m
Long = 0.3 m
Initial temperature = 20°C
Final temperature = 150°C
Suppose the material is copper and here no maintain at time so we assuming heat supplied per unit time
We need to calculate the energy
Using formula of energy
[tex]\Delta Q=mc_{p}\Delta T[/tex]
[tex]\Delta Q=\rho\times V\times c_{p}\times\Delta T[/tex]
[tex]\Delta Q=8960\times\dfrac{\pi}{4}\times(0.015)^2\times0.3\times385\times(423-293)[/tex]
[tex]\Delta Q=23897.6\ J[/tex]
[tex]\Delta Q=23.89\ kJ[/tex]
We need to calculate the power
Using formula of power
[tex]P=\dfrac{\Delta Q}{dt}[/tex]
[tex]P=23.89\ k watt[/tex]
Hence, The maximum power is 23.89 k watt.
A discus thrower accelerates a discus from rest to a speed of 25.3 m/s by whirling it through 1.29 rev. Assume the discus moves on the arc of a circle 0.96 m in radius. A discus thrower moving in a circle as he prepares to throw the discus. (a) Calculate the final angular speed of the discus. rad/s
Answer:
ω = 26.35 rad/s
Explanation:
given,
speed of discus thrower = 25.3 m/s
whirling through = 1.29 rev
radius of circular arc = 0.96 m
final angular speed of the discus = ?
using formula
v = ω r
v is the velocity of disk
ω is the angular speed of the discus
r is the radius of arc
[tex]\omega = \dfrac{v}{r}[/tex]
[tex]\omega = \dfrac{25.3}{0.96}[/tex]
ω = 26.35 rad/s
Final angular speed of the discus is equal to ω = 26.35 rad/s
The final angular speed is approximately 26.354 rad/s. To find the final angular speed, we use the relation [tex]\omega = v / r[/tex]. With v = 25.3 m/s and r = 0.96 m.
Calculate the final angular speed (ω) using the linear speed provided. Rotational motion relies on the relationship between linear speed (v) and angular speed (ω), which is expressed as [tex]v = r\cdot \omega[/tex], where r is the circular path radius. Rearranging this equation yields [tex]\omega = v / r[/tex], demonstrating the angular speed's direct and inverse relationship.
[tex]\omega = v / r[/tex]Here, v = 25.3 m/s (final linear speed) and r = 0.96 m (radius of the circle):
[tex]\omega = 25.3\left m/s / 0.96\left m \approx 26.354\left rad/s[/tex]Conclusion:
Thus, the final angular speed of the discus is approximately 26.354 rad/s.
David is riding on a flying carousel which swings him in circles of radius R=4.0R=4.0m, at a height of h=6h=6 m above the ground. The carousel rotates counterclockwise once every 5 sec.
When he is on the west side of the carousel, David's hat falls off.
Where does the hat land (relative to the point where it fell off)?
Answer:
5.55935324 m
Explanation:
r = Radius = 4 m
h = Height = 6 m
Frequency is
[tex]f=\frac{1}{T}\\\Rightarrow f=\frac{1}{5}\ Hz[/tex]
Angular speed is given by
[tex]\omega=2\pi f\\\Rightarrow \omega=\frac{2}{5}\pi[/tex]
Tangential velocity of the hat is given by
[tex]v=r\omega\\\Rightarrow v=4\times\frac{2}{5}\pi\\\Rightarrow v=5.02654\ m/s[/tex]
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s²
From equation of motion
[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 6=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{6\times 2}{9.81}}\\\Rightarrow t=1.106\ s[/tex]
Distance = Speed×Time
[tex]Distance=5.02654\times 1.106=5.55935324\ m[/tex]
The hat will land 5.55935324 m away from the point of release
David's hat would travel in a straight line relative to the Earth and land to the south of the point where it fell off due to the carousel's counterclockwise rotation and the principles of projectile motion.
Explanation:When we consider David's hat falling off while he is riding on a flying carousel, we have to address the physical principles that govern the motion of the hat after it leaves his head. As David's hat falls off on the west side of the carousel, it would have initially the same horizontal velocity as the carousel at that point. However, since the carousel is rotating counterclockwise, and assuming there is no significant air resistance, the hat would move in a straight line relative to the Earth. This means it would land to the south of the point where it fell off, as the carousel would continue to rotate after the hat has been released. This situation applies principles of projectile motion and inertial frames.
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If you decrease the distance between successive crests of a wave, this changesA. The frequency.B. The wavelength.C. The speed.D. Both the wavelength and the speed.E. Both the wavelength and the frequency
Final answer:
Decreasing the distance between successive crests of a wave results in a shorter wavelength and an increased frequency without affecting the speed of the wave. The correct answer to the student's question is E. Both the wavelength and the frequency.
Explanation:
If you decrease the distance between successive crests of a wave, this changes the wavelength and the frequency of the wave. Decreasing the distance between the crests means the wavelength is shorter, and since wavelength and frequency are inversely related, the frequency increases. However, this does not necessarily change the speed of the wave, which is determined by the medium through which the wave is traveling. Therefore, the correct answer is E. Both the wavelength and the frequency.
Wavelength is defined as the distance between any two consecutive identical points on the waveform, typically measured from crest to crest or trough to trough. When the period of a wave increases, its frequency decreases. This is because the frequency is the number of wave cycles that pass a point in one second, and if the period (the time for one cycle) increases, fewer cycles can pass in the same amount of time.
Two circular plates of radius 9cm are separated in air by 2.0mm, forming a parallel plate capacitor. A battery is connected across the plates. At a particular time, t1, the rate at which the charge is flowing through the battery from one plate to the other is 5A. (a)What is the time rate of change of the electric field between the plates at t1? (b)Compute the displacement current between the plates at t1, and show it is equal to 5A.
Answer:
(a) [tex]2.26\times 10^{13}\ N/C.s[/tex]
(b) 5 A
Solution:
As per the question:
Radius of the circular plate, R = 9 cm = 0.09 m
Distance, d = 2.0 mm = [tex]2.0\times 10^{- 3}\ m[/tex]
At [tex]t_{1}[/tex], current, I = 5 A
Now,
Area, A = [tex]\pi R^{2} = \pi 0.09^{2} = 0.025[/tex]
We know that the capacitance of the parallel plate capacitor, C = [tex]\frac{\epsilon_{o} A}{d}[/tex]
Also,
[tex]q = CV[/tex]
[tex]q = \frac{\epsilon A}{d}V[/tex]
Also,
[tex]V = \frac{E}{d}[/tex]
Now,
(a) The rate of change of electric field:
[tex]\frac{dE}{dt} = \frac{dq}{dt}(\frac{1}{A\epsilon_{o}})[/tex]
where
[tex]I = \frac{dq}{dt} = 5\ A[/tex]
[tex]\frac{dE}{dt} = 5\times (\frac{1}{0.025\times 8.85\times 10^{- 12}}) = 2.26\times 10^{13}\ N/C.s[/tex]
(b) To calculate the displacement current:
[tex]I_{D} = epsilon_{o}\times \frac{d\phi}{dt}[/tex]
where
[tex]\frac{d\phi}{dt}[/tex] = Rate of change of flux
[tex]I_{D} = Aepsilon_{o}\times \frac{dE}{dt}[/tex]
[tex]I_{D} = 0.025\times 8.85\times 10^{- 12}\times 2.26\times 10^{13} = 5\ A[/tex]
The unit of electrical current is the ampere (A).
Which one of the combinations of units below is equivalent to the ampere?
A) C·s
B) C/s
C) N·m/s
D) J·s
E) kg·m²/s
Answer:
option B
Explanation:
Current unit is ampere (A)
Ampere will be equivalent to = ?
we know,
Current can be define as the charge per unit time
[tex]I = \dfrac{Q}{t}[/tex]
unit of charge(Q) is coulomb which is equal to C.
unit of time(t) is equal to 's'.
now,
[tex]I = \dfrac{C}{s}[/tex]
unit of I = C/s
ampere(A) is equivalent to C/s
The correct answer is option B
Final answer:
The unit equivalent to the ampere (A) is C/s (Coulombs per second), making the correct option B) C/s. This illustrates the fundamental relationship between electrical charge, time, and current in physics.
Explanation:
The unit of electrical current is the ampere (A), which is defined as the amount of electrical charge in coulombs that passes through a conductor in one second. The correct unit combination that is equivalent to the ampere is C/s (Coulombs per second). Therefore, the correct option from the given combinations is B) C/s. This unit measures the flow of electrical charge and is a fundamental concept in understanding electric currents and how they are quantified in the International System of Units (SI)
Therefore, as per the above explaination, the correct answer is option C
If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters. View Available Hint(s) f2 = nothing
Answer:
0.84 cm
Explanation:
u = Object distance = 0.35 cm
v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)
f = Focal length
From lens equation
[tex]\frac{1}{f_2}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f_2}=\frac{1}{0.35}+\frac{1}{-0.6}\\\Rightarrow \frac{1}{f_2}=\frac{25}{21}\\\Rightarrow f_2=\frac{21}{25}=0.84\ cm[/tex]
Focal length of the lens is 0.84 cm
A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling. The block swings in an arc, rising 3 mm above its lowest position. The velocity of the bullet was:
Answer:
Velocity of the bullet was 800 m/s.
Explanation:
It is given that,
Mass of the block, m₁ = 10 kg
Mass of the bullet, m₂ = 3 g
Height reached by the block, h = 3 mm
Let v is the velocity of the wooden block. Using the conservation of energy to find it as :
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 3\times 10^{-3}}[/tex]
v = 0.24 m/s
Let v' is the velocity of the bullet. The momentum of the system remains conserved. It can be calculated as :
[tex]m_1v=m_2v'[/tex]
[tex]v'=\dfrac{m_1v}{m_2}[/tex]
[tex]v'=\dfrac{10\times 0.24}{3\times 10^{-3}}[/tex]
v' = 800 m/s
So, the velocity of the bullet was 800 m/s. Hence, this is the required solution.
The velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.
What is conservation of momentum?When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling.
The kinetic energy will be equal to the gravitation potential energy for this case. Thus,
[tex]\dfrac{1}{2}mv^2=mgh\\v^2=2gh[/tex]
As the maximum height is 3mm or 0.003 m. Thus, the velocity of the wooden block is,
[tex]v^2=2(9.8)(0.003)\\v=0.24\rm\; m/s[/tex]
Suppose the velocity of the bullet is v'. Thus by the law of conservation of momentum,
[tex]m_1v=m_2v'\\10(0.24)=0.003(v')\\v'=800\rm\; m/s[/tex]
Hence, the velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.
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A 2.15-MHz sound wave travels through a pregnant woman’s abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 90 beats per second are detected. The speed of sound in body tissue is 1510 m/s. Calculate the speed of the fetal heart wall at the instant this measurement is made.
Answer:
v= 0.0316 m/s
Explanation:
We need to use the Doppler Effect defined as the change in frequency of a wave in relation to an observer who is moving relative to the wave source.
Notation
Let v= magnitude of the heart wall speed
V= speed of sound
fh= the frequency the heart receives (and reflects)
fi= original frequency
ff= reflected frequency
fb= frequency for the beats
Apply the Doppler Effect formula
Since the heart is moving observer then the device is a stationary source, and we have this formula
fh = [(V+ v)/(v)] fi (1)
We can consider the heart as moving source and the device as a stationary observer, and we have this formula
ff = [(V)/(V-v)] fh (2)
The frequency for the beats would be the difference from the original and the reflected frequency
fb = ff -fi (3)
Replacing equations (1) and (2) into equation (3) we have:
[tex] f_b = \frac{V}{V-v} \frac{V+v}{V}f_i - f_i [/tex]
[tex]f_b = f_i ( \frac{V+v}{V-v} -1) [/tex]
fb = fi(V+v -V+v)/(V-v)
[tex] f_b = \frac{2v}{V-v}[/tex]
Solving for v we have:
[tex] v = V (\frac{f_b}{2f_o - f_b}) [/tex]
[tex] v = 1510 m/s (\frac{90 Hz}{2∗2150000Hz - 90 Hz})= 0.0316 m/s [/tex]
The speed of the fetal heart wall at the instant this measurement is made can be calculated using the Doppler shift in frequency. Given the transmitted sound frequency of 2.15 MHz and beat frequency detected as 90 Hz, the speed of the heart wall can be calculated using the provided speed of sound in body tissue. The calculated speed is approximately 0.063 m/s.
Explanation:The question is based on the application of the principle of the Doppler effect in medical physics, specifically radiology. The Doppler effect involves a change in frequency and wavelength of a wave in relation to an observer who is moving relative to the wave source. In the case of ultrasound waves being used to monitor a fetus in the womb, the opening and closing of the heart valves reflect the waves back, and by calculating the shift in frequency (the Doppler shift), we can find out the speed at which the heart wall is moving.
In this particular scenario, two major factors are at play, namely the frequency of the transmitted sound (2.15 MHz), and the beat frequency detected (90 Hz). To calculate the speed of the fetal heart wall at the instant this measurement is made, we use the formula for Doppler shift in frequency: Δf/f=V/Vw, where Δf is the change in frequency (i.e., the beat frequency), V is the velocity of the moving observer (i.e., the fetal heart wall), and Vw is the speed of sound in the medium (body tissue in this case, which is given as 1510 m/s).
So our calculation will be as follows: V=(Δf/f) * Vw. Plugging in our values, V = (90 s-1 / 2.15 x 106 s-1) * 1510 m/s, which yields a value for V, the speed of the fetal heart wall, as approximately 0.063 m/s.
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Which one of the following statements concerning kinetic energy is true? a The kinetic energy of an object always has a positive value. b The kinetic energy of an object is directly proportional to its speed. c The kinetic energy of an object is expressed in watts. d The kinetic energy of an object is a quantitative measure of its inertia. e The kinetic energy of an object is always equal to the object’s potential energy.
a The kinetic energy of an object always has a positive value.
Explanation:
The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
Let's now analyze each statement:
a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term [tex]v^2[/tex] is always positive as well, so the kinetic energy is always positive.
b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, [tex]K\propto v^2[/tex].
c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.
d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.
e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.
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Kinetic energy, a concept in physics, refers to the energy that an object possesses due to its motion and is always positive. It is proportional to the square of speed, expressed in Joules, and distinct from inertia and potential energy.
Explanation:In physics, kinetic energy pertains to the energy possessed by an object due to its motion. The kinetic energy (K) of a moving object can be calculated using the formula K = (1/2)mv², where 'm' stands for mass and 'v' for velocity.
Concerning the statements included in the question, option a) 'The kinetic energy of an object always has a positive value.' is correct. This is due to the fact that kinetic energy is product of the mass and the square of the speed, which are always positive or zero, thereby making kinetic energy always positive.
Regarding option b) 'The kinetic energy of an object is directly proportional to its speed', it's essential to note that kinetic energy is proportional not to the speed, but the square of the speed, hence this statement is partially correct.
The kinetic energy is not expressed in watts (option c). It's actually measured in Joules, which is equal to kg · m²/s² (kilogram meter squared per second squared).
Kinetic energy is not a measure of an object’s inertia (option d). It measures an object's energy due to motion, whereas inertia is an object’s resistance to change in motion.
Lastly, the kinetic energy of an object is not always equal to the object’s potential energy (option e). These are two different forms of energy that can convert into each other under certain circumstances, but they are not always equal.
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When light with a frequency f1 = 547.5 THz illuminates a metal surface, the most energetic photoelectrons have 1.260 x 10^-19 J of kinetic energy. When light with a frequency f2 = 738.8 THz is used instead, the most energetic photo-electrons have 2.480 x 10^-19 J of kinetic energy
Using these experimental results, determine the approximate value of Planck's constant.
Answer:
The approximate value of Planck's constant is [tex]6.377\times10^{-34}\ J[/tex]
Explanation:
Given that,
Frequency [tex]f_{1}= 547.5\ THz[/tex]
Kinetic energy [tex]K.E=1.260\times10^{-19}\ J[/tex]
Frequency [tex]f_{2}=738.8\ THz[/tex]
Kinetic energy [tex]K.E=2.480\times10^{-19}\ J[/tex]
We need to calculate the approximate value of Planck's constant
Using formula of change in energy
[tex]E = hf[/tex]
[tex]K.E_{2}-K.E_{1}=h(f_{2}-f_{1})[/tex]
[tex]h=\dfrac{K.E_{2}-K.E_{1}}{(f_{2}-f_{1})}[/tex]
[tex]h=\dfrac{2.480\times10^{-19}-1.260\times10^{-19}}{738.8\times10^{12}-547.5\times10^{12}}[/tex]
[tex]h=6.377\times10^{-34}\ J[/tex]
Hence, The approximate value of Planck's constant is [tex]6.377\times10^{-34}\ J[/tex]
A 50-N crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50. A 20-N force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?
a. 20 N to the left.
b. 25 N to the left.
c. 25 N to the right.
d. None of the above; the crate starts to move.
e. 20 N to the right.
Answer:
b. 25 N to the left.
Explanation:
Hi there!
The static friction force is calculated as follows:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of static friction.
If the object is not being accelerated in the vertical direction, the normal force is equal to the weight of crate (with opposite sign). Then:
Fr = 50 N · 0.50
Fr = 25 N
Since the 20-N force is applied to the right of the box, the friction force will be directed to the left because the friction force opposes the displacement. Then, the right answer is "b": 25 N to the left.
The resulting static friction force on the crate will be 20 N to the left, matching the applied force but acting in the opposite direction to prevent the crate from moving.
Explanation:The question is related to static and kinetic friction in physics. A 50-N crate on a horizontal floor has a coefficient of static friction (μs) of 0.50. When a 20-N force is applied, the force of static friction (fs) that acts on the crate in response can be calculated. Since the applied force is less than the maximum possible static friction force (μs × normal force), the static friction will equal the applied force but act in the opposite direction, preventing the crate from moving. Here, the normal force (N) is equal to the weight of the crate because the floor is horizontal and there is no vertical acceleration.
The maximum static friction force would be fs(max) = μs × N = 0.50 × 50 N = 25 N. However, the applied force is only 20 N. Since this is less than the maximum, the actual static frictional force will match the applied force of 20 N, opposing the direction of the applied force, which is to the left. Thus, the correct answer is a. 20 N to the left.
The rear wheels of a truck support 57.0 % of the weight of the truck, while the front wheels support 43.0 % of the weight. The center of gravity of the truck is 1.68 m in front of the rear wheels.
What is the wheelbase of the truck (the distance between the front and rear wheels)? Express your answer with the appropriate units.
To solve the problem it is necessary to apply the concepts related to torque, as well as the concepts where the Force is defined as a function of mass and acceleration, which in this case is gravity.
Considering the system in equilibrium, we perform sum of moments in the rear wheel (R2)
[tex]\sum M = 0[/tex]
[tex]F_g*1.68-R_1*d = 0[/tex]
[tex]mg*1.68-R_1*d = 0[/tex]
Another of the parameters given in the problem is that the front wheel supports 43% of the weight, that is
[tex]R1=0.43F_g[/tex]
[tex]R1=0.43mg[/tex]
Replacing, we have to
[tex]mg*1.68 -R_1*d = 0[/tex]
[tex]mg*1.68 -0.43mg*d = 0[/tex]
[tex]mg*1.68 =0.43mg*d[/tex]
[tex]1.68 = 0.43*d[/tex]
[tex]d =3.9m[/tex]
Therefore the wheelbase of the truck is 3.9m between the front and the rear.
With the use of physics principles, the truck's wheelbase, given that the rear wheels support 57.0% of the weight, and this weight acts at the center of gravity 1.68 meters in front of the rear wheels, is found to be 3.90 meters.
Explanation:The question asks for the wheelbase of the truck, which can be solved using physics principles such as torque and equilibrium. The center of gravity is the point where all the weight can be considered to be concentrated for the purpose of calculations. Here, we know that the rear wheels support 57.0% of the weight of the truck, and this weight acts at the center of gravity, which is 1.68 meters in front of the rear wheels. Given that the truck is in equilibrium (i.e., not tipping over), the torques about any point caused by the weight must cancel out. Hence, the total distance from the rear wheels (where the 57% weight acts) to the front wheels (where the 43% weight acts) is (1.68 m * 57.0 / 43.0) m = 2.22 meters. Therefore, the wheelbase of the truck is 1.68 m + 2.22 m = 3.90 meters.
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A 4.00 m-long steel cable is used to support a church chandelier. After the 226-kg chandelier is hung from the ceiling, the steel cable is lengthened by 3.00 mm. Calculate the diameter of the cable in millimeters. (The Young’s modulus for steel is ????????. ???????????????? × ???????????????????????????????? ????????/????????????????.)
Answer:
4.22 mm
Explanation:
E = Young’s modulus for steel = 210 GPa (generally)
[tex]\Delta L[/tex] = Change in length = 3 mm
[tex]L_0[/tex] = Original length = 4 m
A = Area of cable
g = Acceleration due to gravity = 9.81 m/s²
r = Radius of cable
d = Diameter = 2r
m = Mass of chandelier = 226 kg
[tex]\epsilon[/tex] = Longitudinal strain = [tex]\frac{\Delta L}{L_0}[/tex]
Uniaxial stress is given by
[tex]\sigma=E\epsilon\\\Rightarrow \sigma=210\times 10^9 \frac{3\times 10^{-3}}{4}\\\Rightarrow \sigma=157500000\ Pa[/tex]
[tex]\sigma=\frac{F}{A}\\\Rightarrow \sigma=\frac{mg}{\pi r^2}\\\Rightarrow 157500000=\frac{226\times 9.81}{\pi r^2}\\\Rightarrow r=\sqrt{\frac{226\times 9.81}{157500000\times \pi}}\\\Rightarrow r=0.00211\ m\\\Rightarrow d=2r\\\Rightarrow d=2\times 2.11=4.22\ mm[/tex]
The diameter of the cable is 4.22 mm
You are trying to determine the specific gravity of an unknown liquid. If m is the mass of a solid object, mL is the object's apparent mass in the unknown liquid, and mA is the apparent mass of the object in water, what is the formula for the specific gravity of the liquid?
Final answer:
The specific gravity of a liquid can be calculated using the formula: Specific Gravity = mL / mA. Specific gravity is a dimensionless number used to compare the density of a substance with the density of water.
Explanation:
The specific gravity of a liquid can be calculated using the formula:
Specific Gravity = mL / mA
Here, mL is the apparent mass of the object in the unknown liquid, and mA is the apparent mass of the object in water. Specific gravity is a dimensionless number, so it doesn't have any units. It is used to compare the density of a substance with the density of water, which is 1.0 g/mL.
A solid disk of mass 3 kg, radius 0.5 m, and initial angular velocity 10 rad/sec is dropped onto an initially stationary second disk having the same radius and twice the mass. The second disk cannot fall, but is free to rotate. The disks stick and rotate together. Assume that there are no external torques. Please find each of the following.
a) The kinetic energy of the first disk before it is dropped
b) The final angular velocity of the combined disks
c) The final kinetic energy of the combined disks
The first disk's kinetic energy before it is dropped is 18.75 J. The final angular velocity of the combined disks after they stick together is 3.33 rad/s. The final kinetic energy of the combined disks is 24.97 J.
Explanation:Given a solid disk of mass 3 kg, radius 0.5 m, and initial angular velocity 10 rad/sec, and a second disk with the same radius but with double the mass (6 kg), which is initially at rest, we need to find the following:
Kinetic energy of the first disk before it is droppedFinal angular velocity of the combined disks after they stick togetherFinal kinetic energy of the combined disks(a) The kinetic energy (KE) of the first disk before it is dropped is calculated using the formula for rotational kinetic energy, KE = (1/2)Iω², where I is the moment of inertia for a solid disk (I = (1/2)mr²) and ω is the angular velocity. Therefore, KE = (1/2)((1/2)mr²)ω² = (1/2)((1/2)(3 kg)(0.5 m)²)(10 rad/s)² = 18.75 J.
(b) To find the final angular velocity (ω'), we conserve angular momentum since no external torques are acting. L_initial = L_final, which means Iω = (I1 + I2)ω'. Calculating, we get (1/2)(3 kg)(0.5 m)²(10 rad/s) = ((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)ω', solving for ω' gives us 3.33 rad/s.
(c) The final kinetic energy (KE_final) is calculated with the combined moment of inertia and the final angular velocity using KE_final = (1/2)(I1 + I2)ω'^2. Using the values from (b), we have KE_final = (1/2)((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)(3.33 rad/s)² = 24.97 J.
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W per channel. Part A In terms of dB, how much louder will the more powerful amplifier be when both are producing sound at their maximum levels?
Answer:
The sound level will be 1.870 dB louder.
Explanation:
Given that,
Power = 130 W
Power = 200 W
We need to calculate the sound level
Using formula of sound level
[tex]I_{dB}=10\log(\dfrac{I}{I_{0}})[/tex]
For one amplifier,
[tex]I_{1}=10\log(\dfrac{130}{I_{0}})[/tex]...(I)
For other amplifier,
[tex]I_{2}=10\log(\dfrac{200}{I_{0}})[/tex]...(II)
For difference in dB levels
[tex]I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})[/tex]
[tex]I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})[/tex]
[tex]I_{2}-I_{1}=10\log(\dfrac{200}{130})[/tex]
[tex]I_{2}-I_{1}=1.870\ dB[/tex]
Hence, The sound level will be 1.870 dB louder.
A new experiment is created with the screen at a distance of 1.9 m from the slits (with spacing 0.11 mm). What is the distance between the second order bright fringe of light with λ = 695 nm and the third order bright fringe of light with λ = 407 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.) |x| =
Answer:
The distance between the bright fringes is 2.19 mm
Explanation:
Distance of the screen from the slits, d = 1.9 m
Slit width, w = 0.11 mm = [tex]0.11\times 10^{- 3}\ m[/tex]
Wavelength, [tex]\lambda = 695\ nm[/tex]
Wavelength, [tex]\lambda' = 407\ nm[/tex]
To calculate the distance between the second order bright fringe:
[tex]y_{n} = \frac{n\lambda d}{w}[/tex]
[tex]y'_{3} = \frac{n\lambda' d}{w} = \frac{3\times 695\times 10^{- 9}\times 1.9}{0.11\times 10^{- 3}} = 0.036\ m[/tex]
[tex]y_{2} = \frac{2\times 407\times 10^{- 9}\times 1.9}{0.11\times 10^{- 3}} = 0.014\ m[/tex]
Distance, |x| = [tex]y'_{3} - y_{2}[/tex]
|x| = [tex]0.036 - 0.0141 = 0.0219 m = 2.19\ mm[/tex]
A city uses a water tower to store water for times of high demand. When demand is light, water is pumped into the tower. When demand is heavy, water can flow from the tower with- out overwhelming the pumps. To provide water at a typical 350 kPa gauge pressure, how tall must the tower be?
Answer:
The height of the tower will be 35.714 m
Explanation:
We have given gauge pressure [tex]P=350kPa=350\times 10^3Pa[/tex]
Density of water [tex]\rho =1000kg/m^3[/tex]
We have to find the height of the tower h
We know that gauge pressure is given by [tex]P=\rho gh[/tex]
[tex]350\times 10^3=100 0\times 9.8\times h[/tex]
[tex]h=35.714m[/tex]
So the height of the tower will be 35.714 m
The height of the water tower needs to be determined in order to provide water at a 350 kPa gauge pressure. By using the equation for hydrostatic pressure and the given values, the height is calculated to be 35.71 meters.
Explanation:To provide water at a typical 350 kPa gauge pressure, the height of the water tower needs to be determined. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, and it is calculated with the equation p = hpg, where p is the pressure, h is the height, and ρ is the density of the fluid.
In this case, the fluid is water and the density of water is 1000 kg/m³. Rearranging the equation, we can solve for h: h = p / (ρg). Using the given pressure of 350 kPa and the acceleration due to gravity of 9.8 m/s², we can calculate the height of the water tower,
h = 350,000 Pa / (1000 kg/m³ × 9.8 m/s²) = 35.71 m
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7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surface, at what depth will it be down to 1 μV/m? Water has μr = 1, and at 1 GHz, r = 80 and σ = 1 S/m.
The depth at which the amplitude of the incident wave is down to 1 μV/m in water can be calculated using the equations for wave propagation. At 1 GHz, the skin depth of water is approximately 5 mm. Using the exponential attenuation equation, we can find that the depth is approximately 35.3 mm.
Explanation:To find the depth at which the amplitude of the incident wave is down to 1 μV/m, we can use the equations for wave propagation in a medium.
First, we need to calculate the skin depth (δ) of the water at 1 GHz using the equation:
δ = √(2/πfμσ)
Substituting the given values, we get:
δ = √(2/π * 1 * 10^9 * 4π * 10^-7 * 1) = 0.005 m = 5 mm
Next, we can use the equation for exponential attenuation of the wave:
A = A0 * e^(-x/δ)
Here, A0 is the initial amplitude, A is the amplitude at depth x, and δ is the skin depth. We can rearrange the equation to solve for x:
x = -δ * ln(A/A0)
Substituting the given values, we get:
x = -5 mm * ln(1 * 10^-6/20) ≈ 35.3 mm
Therefore, the depth at which the amplitude of the incident wave is down to 1 μV/m is approximately 35.3 mm.
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Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg object is at the origin, and a 3.99-kg object is at -0.496 m.
Where is the center of mass of these objects?
Answer:
0.958 m
Explanation:
So the total mass of the system is
M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28 kg
let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation
[tex]My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4[/tex]
[tex]11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806[/tex]
[tex]y = \frac{10.806}{11.28} = 0.958 m[/tex]
So the center of mass is 0.958 m from the origin
One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 810 km above Earth's surface collides with a pellet having mass 3.7 g. (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a 3.7 g bullet from a modern army rifle with a muzzle speed of 1200 m/s?
Answer:
411087.52089 J
[tex]\frac{K_r}{K_b}=154.31213[/tex]
Explanation:
R = Radius of Earth = 6370000 m
h = Altitude of satellite = 810 km
r = R+h = 63700000+810000 m
m = Mass of bullet = 3.7 g
Velocity of bullet = 1200 m/s
The relative velocity between the pellets and satellite is 2v
Now, the square of velocity is proportional to the kinetic energy
[tex]K\propto v^2[/tex]
[tex]\\\Rightarrow 4K\propto (2v)^2\\\Rightarrow 4K\propto 4v^2[/tex]
Kinetic energy in terms of orbital mechanics is
[tex]K=\frac{GMm}{2r}[/tex]
In this case relative kinetic energy is
[tex]K_r=4\frac{GMm}{2r}\\\Rightarrow K_r=2\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}\times 3.7\times 10^{-3}}{(6370+810)\times 10^3}\\\Rightarrow K_r=411087.52089\ J[/tex]
The relative kinetic energy is 411087.52089 J
The ratio of kinetic energies is given by
[tex]\frac{K_r}{K_b}=\frac{411087.52089}{\frac{1}{2}\times 3.7\times 10^{-3}\times 1200^2}\\\Rightarrow \frac{K_r}{K_b}=154.31213[/tex]
The ratio is [tex]\frac{K_r}{K_b}=154.31213[/tex]