Answer:
W = 1.49 10⁻¹¹ kg
Explanation:
For this problem, let's use Newton's equation of equilibrium
F - W = 0
F = W (1)
Strength can be found from the definition of pressure
P = F / A
F = P A
The radiation pressure for a reflective surface is
P = 2 I / c)
We substitute in equation 1
2 I / c A = W
The intensity is defined by the ratio of the power between the area
I = P / A
P = I A
We substitute
2 P / c = W
W = 2 2.24 10-3 / 3 108
W = 1.49 10⁻¹¹ kg
Two monatomic gases, helium and neon, are mixed in a sealed container and brought into thermal equilibrium at temperature T . If the molar mass of helium is 4.0 g/mol and the molar mass of neon is 20.2 g/mol , then _______.
A. all the atoms have the same average speed
B. the average speed of the neon atoms is greater than the average speed of the helium atoms
C. the average speed of the helium atoms is greater than the average speed of the neon atoms
D. the atoms diffuse from high temperature to low temperature
E. all the atoms have exactly the same velocity.
In a mixture of helium and neon gases at thermal equilibrium, the average speed of helium atoms will be greater than the average speed of neon atoms.
Explanation:In a mixture of two gases at thermal equilibrium, the average speed of the atoms will depend on their respective molar masses. The average speed of an atom can be calculated using the root mean square speed formula, v = sqrt(3kT/m), where v is the average speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the atom. In this case, helium has a smaller molar mass than neon, so according to the formula, the average speed of helium atoms will be greater than the average speed of neon atoms.
Learn more about Average speed here:https://brainly.com/question/12322912
#SPJ12
The average speed of helium atoms is higher than that of neon atoms because the molar mass of helium is much smaller than that of neon. At the same temperature, lighter molecules move faster. Thus, the correct answer is C.
To determine the average speeds of helium and neon atoms in a mixture at thermal equilibrium at temperature T, we need to consider the relationship between temperature, mass, and molecular speed.
The average kinetic energy per atom at a given temperature T is the same for both helium and neon. This can be expressed as:
[tex]\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T[/tex]
where,
[tex]k_B[/tex]= Boltzmann constant
[tex]T[/tex]= temperature
[tex]m[/tex]= molar mass of the gas
Since both gases have the same average kinetic energy, the differences in their masses will influence their average speeds. Specifically, the equation for the root-mean-square speed (v_{rms}) of a gas is:
[tex]v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}[/tex]
Given:
Molar mass of helium (He) =[tex]4.0 g/mol[/tex]
Molar mass of neon (Ne) = [tex]20.2 g/mol[/tex]
We see that helium has a much smaller molar mass compared to neon. Therefore, for the same temperature T:
[tex]v_{rms, He} > v_{rms, Ne}[/tex]
This means the average speed of helium atoms is greater than that of neon atoms. Thus, the correct answer is:
C. the average speed of the helium atoms is greater than the average speed of the neon atoms.
Problem 3: An Nd:YAG laser operates in a pulsed mode, with an energy of 100mJ (millimoles) per pulse, and a pulse repetition rate of 10Hz. Light can be emitted at 1065 nm, 532 nm, or 355 nm. (a) Each pulse lasts for 1ns, and in between pulses, no light is emitted. What is the instantaneous laser power during each pulse
Given Information:
Energy of laser pulses = E = 100 mJ = 100×10⁻³ Joules
Time = t = 1 ns = 1×10⁻⁹ seconds
Required Information:
Instantaneous power = P = ?
Answer:
[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]
Explanation:
The instantaneous power is the power dissipated at any instant of time whereas the average power is the power dissipated over a given time interval.
The instantaneous laser power during each pulse is given by
[tex]Instantaneous \: Power = \frac{E}{t}[/tex]
Where E is the energy of the laser pulses and t is the time that each pulse lasts.
[tex]Instantaneous \: Power = \frac{100\times10^{-3}}{1\times10^{-9}}[/tex]
[tex]Instantaneous \: Power = 1\times10^{8} \: watt[/tex]
or
[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]
Therefore, the instantaneous power of each pulse is 100 Mwatt.
The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they act like a single solenoid?
Answer: 13.1 μH
Explanation:
Given
length of heating coil, l = 1 m
Diameter of heating coil, d = 0.8 cm = 8*10^-3 m
No of loops, N = 400
L = μN²A / l
where
μ = 4π*10^-7 = 1.26*10^-6 T
A = πd²/4 = (π * .008 * .008) / 4 = 6.4*10^-5 m²
L = μN²A / l
L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1
L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5
L = 1.31*10^-5
L = 13.1 μH
Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH
Answer:
Their total self-inductance assuming they act like a single solenoid is 10.11 μH
Explanation:
Given;
diameter of the heating coil, d = 0.800 cm
combined length of heating coil and hair dryer, [tex]l[/tex] = 1.0 m
number of turns, N = 400 turns
Formula for self-inductance is given as;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where
μ₀ is constant = 4π x 10⁻⁷ T.m/A
A is the area of the coil:
A = πd²/4
A = π (0.8 x 10⁻²)²/4
A = 5.027 x 10⁻⁵ m²
[tex]L = \frac{\mu_oN^2 A}{l } = \frac{4\pi *10^{-7}(400)^2 *5.027*10^{-5}}{1 } \\\\L =1.011 *10^{-5} \ H\\\\L = 10.11 \mu H[/tex]
Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH
Two slits separated by a distance of d = 0.12 mm are located at a distance of D = 0.63 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of λ = 540 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. What is the pathlength difference between the waves at the second maximum (m=2) on the screen?
Answer:
The path-length difference is [tex]dsin\theta=1.08*10^{-3}mm[/tex]
The angle is [tex]\theta = 0.5157^o[/tex]
Explanation:
From the question we are told that
The distance of separation is d = 0.12 mm = [tex]0.12*10^{-3} m[/tex]
The distance from the screen is D = 0.63 m
The wavelength is [tex]\lambda = 540nm = 540 *10^{-9}m[/tex]
From the question we can deduce that the the two maxima's are at the m=0 and m=2
Now the path difference for this second maxima is mathematically represented as
[tex]d sin \theta = m \lambda[/tex]
Where d[tex]dsin\theta[/tex] is the path length difference
Substituting values
[tex]dsin \theta = 2 * 540*10^{-9}[/tex]
[tex]dsin\theta = 1.08*10^{-6}m[/tex]
converting to mm
[tex]dsin\theta = 1.08*10^{-6} * 1000 mm[/tex]
[tex]dsin\theta=1.08*10^{-3}mm[/tex]
To obtain the angle we make [tex]\theta[/tex] the subject
[tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
Substituting values
[tex]\theta = sin ^{-1} [\frac{1.08*10^{-6}}{0.12*10^-3} ][/tex]
[tex]\theta = 0.5157^o[/tex]
The Pathlength difference between the waves at second maximum on the screen is; 1.08 × 10^(-6) m
We are given;
Distance between two slits; d = 0.12 mm = 0.12 × 10^(-3) m
Distance of slit from screen; D = 0.6 m
We want to find the path length at second maxima m = 2
λ = 540 nm = 540 × 10^(-9) m
Formula for Pathlength is;
dsin θ = mλ
Where mλ is the Pathlength difference.
Since at m = 0, the pathlength is zero,
Thus;
Pathlength difference = (2 × 540 × 10^(-9)) - 0
Pathlength difference = 1.08 × 10^(-6) m
Read more at; https://brainly.com/question/12645865
Consider an experiment in which slow neutrons of momentum ¯hk are scattered by a diatomic molecule; suppose that the molecule is aligned along the y axis, with one atom at y = b and the other at y = −b. The beam of neutrons is directed in the zb direction. Assume the atoms to be infinitely heavy so that they remain fixed throughout the experiment. The potential due to the atoms as seen by the neutrons can be represented by a delta function, so: V (~r) = a[δ(x)δ(y − b)δ(z) + δ(x)δ(y + b)δ(z)] (a) Calculate the scattering amplitude, and differential cross section, in the first order Born approximation. (b) In what ways does the quantum result differ from what one would expect classically?
Answer:
Check the explanation
Explanation:
When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.
position = amplitude x sine function(angular frequency x time + phase difference)
x = A sin(ωt + ϕ)
x = displacement (m)
A = amplitude (m)
ω = angular frequency (radians/s)
t = time (s)
ϕ = phase shift (radians)
Kindly check the attached image below to see the step by step explanation to the question above.
Problem 24.3 The assembly is made from a steel hemisphere, rho st = 7. 80 Mg/m3 , and an aluminum cylinder, rho al = 2. 70 Mg/m3 . If the height of the cylinder is h = 180 mm, determine the location z of the mass center of the assembly.
Given:
[tex]\rho_s = 7.8 \ Mg/m^3[/tex][tex]\rho_a = 2.70 \ Mg/m^3[/tex]Height of cylinder, [tex]h = 180 \ mm[/tex]Now,
The mass of the steel hemisphere will be:
→ [tex]m_h = \rho_s\times \frac{2}{3} \pi r^2[/tex]
By putting the values, we get
[tex]= 7.8\times 10^3\times \frac{2}{3} \pi (0.16)^2[/tex]
[tex]= 66.9 \ kg[/tex]
and,
The mass of aluminum cylinder will be:
→ [tex]m_c = \rho_a \times \pi r^2 h[/tex]
[tex]= 2.7\times 10^3\times \pi (0.08)^2 (0.18)[/tex]
[tex]= 9.7 \ kg[/tex]
Now,
The mass center of steel hemisphere will be:
→ [tex]z_1 = r - \frac{3r}{8}[/tex]
[tex]= 0.16-(3\times \frac{0.16}{8} )[/tex]
[tex]= 0.1 \ m[/tex]
The mass center of aluminum hemisphere will be:
→ [tex]z_2 = r+\frac{h}{2}[/tex]
[tex]= 0.16+\frac{0.18}{2}[/tex]
[tex]= 0.25 \ m[/tex]
hence,
The center of mass will be:
→ [tex]Z = \frac{m_h z_1 +z_2 m_c}{m_s+m_c}[/tex]
By putting the values, we get
[tex]= \frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76}[/tex]
[tex]= 0.12 \ m[/tex]
Thus the above answer is right.
Learn more about mass center here:
https://brainly.com/question/14931324
To find the location (z) of the mass center of a steel hemisphere and an aluminum cylinder assembly, we first calculate the mass of each part using the given densities and geometric properties. We then use a formula to calculate 'z' based on the mass and z-coordinate of the center of mass of each part. The center of mass assumes the symmetric mass distributions.
Explanation:To determine the location z of the mass center of the assembly, we first need to calculate the mass of each part of the assembly using their given densities and geometrical properties.
The mass of the steel hemisphere (mSt) is calculated as density times the volume. The volume of the hemisphere can be calculated using the formula 2/3*Pi*r^3, where r is the radius of the hemisphere.
The mass of the aluminum cylinder (mAl) is calculated the same way, as the product of the cylinder's density and volume. The volume of a cylinder is given by the formula Pi*r^2*h, where r is the radius and h is the height of the cylinder, which is given as 180 mm or 0.12 m.
Once we have these masses, we then calculate the mass center z of the assembly, which is given by the formula z = (mSt * zSt + mAl * zAl) / (mSt + mAl), where zSt and zAl are the z-coordinates of the center of the mass of the steel hemisphere and the aluminum cylinder respectively.
Since the mass of each body acts as if it were at the center of the body, we can assume that the z-coordinate of the hemisphere (zSt) is at its geometric center, whereas for the cylinder (zAl), it would be at h/2 from the bottom end.
Learn more about Center of Mass here:https://brainly.com/question/33607198
#SPJ11
A common technique in analysis of scientific data is normalization. The purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. The goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. In this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. Suppose that in Part 2.2.2 of the experiment, students obtain a signal value of 185 mV at a distance of 4 cm and a value of 82 mV at a distance of 6.2 cm. Normalize the students' data to the value obtained at 4 cm. (Divide the signal value by 185.) Then calculate the theoretically expected (normalized) value at 6.2 cm.___________________ Normalized experimental value at 6.2 cm____________________Theoretically expected normalized value at 6.2 cm
Answer:
1. Normalized value at 6.2 cm = 0.443
2. Theoretical expected value = 0.416
Explanation:
1. Normalized experimental value is calculated as follows;
Normalized experimental value = 82/185
= 0.443
Therefore, normalized value at 6.2 cm = 0.443
2. Calculating the theoretical expected value using the relation;
V₁r₁² = V₂r₂²
V₂ = V₁(r₁/r₂)²
= 1* (4/6.2)²
= 1 *0.645²
= 0.416
Therefore, the theoretical expected value = 0.416
After watching a news story about a fire in a high rise apartment building, you and friend decide to design an emergency escape device from the top of a building. To avoid engine failure, your friend suggest a gravitational powered elevator. The design has a large, heavy turntable (a horizontal disk that is free to rotate about its center) on teh roof with a cable wound around its edge. The free end of the cable goes horizontally to the edge of the building roof, pases over a heavy vertical pulley, and then hangs straight down. A strong wire cage which can hold 5 people is then attached to the hanging end of the cable. When people enter the cage adn release it, the cable unrolls from the turntable lowering the people as safely to the ground. to see if the design is feasible you decide to calculate the acceleration of the fully loaded elevator to make sure it is much less than g. Your friend design has the radius of the turntable disk as 1.5 m and its mass as 2x that of the fully loaded elevator. The disk which serves as the vertical pulley has 1/4 the radius of the turntable and 1/16 its mass. In your physics book you find that the moment of inertia if a disk is 1/2 that of a ring.
Answer:
You have been hired as part of a research team consisting of ... Torque: After watching a news story about a fire in a high rise apartment building, you and your friend decide to design an emergency escape device from the top ... To avoid engine failure, your friend suggests a gravitational powered elevator.
Explanation:
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at v i , 2 = 17.3 vi,2=17.3 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle θ = 39.9 θ=39.9 ° below the horizontal and a speed of v i , 1 = 33.7 vi,1=33.7 m/s. What is the speed of the eagle immediately after it catches its prey?
The speed of the eagle immediately after catching the pigeon is [tex]$19.0$[/tex] m/s.
To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction.
Given information:
- Mass of the pigeon = [tex]$m_2$[/tex]
- Mass of the eagle = [tex]$2m_2$[/tex] (twice the mass of the pigeon)
- Initial velocity of the pigeon, [tex]$\vec{v}_{i,2} = 17.3 m/s[/tex] (northward)
- Initial velocity of the eagle, [tex]$\vec{v}_{i,1} = 33.7 m/s[/tex] at an angle [tex]$\theta = 39.9^\circ$[/tex] below the horizontal
We need to find the final velocity of the eagle after catching the pigeon, [tex]$\vec{v}_f$[/tex].
Step 1: Resolve the initial velocity vectors into components.
For the pigeon:
[tex]$\vec{v}_{i,2} = (0, 17.3) m/s[/tex]
For the eagle:
[tex]$\vec{v}_{i,1} = (33.7 \cos 39.9^\circ, -33.7 \sin 39.9^\circ) m/s[/tex]
[tex]$\vec{v}_{i,1} = (22.8, -25.4) m/s[/tex]
Step 2: Calculate the initial total momentum of the system.
Initial total momentum, [tex]$\vec{p}_i = m_2 \vec{v}_{i,2} + (2m_2) \vec{v}_{i,1}$[/tex]
[tex]$\vec{p}_i = m_2 (0, 17.3) + 2m_2 (22.8, -25.4)$[/tex]
[tex]$\vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]
Step 3: Calculate the final total momentum of the system.
Since momentum is conserved, the final total momentum is equal to the initial total momentum.
[tex]$\vec{p}_f = \vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]
Step 4: Calculate the final velocity of the combined system (eagle + pigeon).
Let [tex]$\vec{v}_f = (v_{f,x}, v_{f,y})$[/tex] be the final velocity of the combined system.
Total final momentum = [tex]$(m_2 + 2m_2) \vec{v}_f = 3m_2 \vec{v}_f$[/tex]
[tex]3m_2 \vec{v}_f = m_2 (45.6, -33.8)$$\vec{v}_f = \left(\frac{45.6}{3}, -\frac{33.8}{3}\right)$ m/s\\$\vec{v}_f = (15.2, -11.3)$ m/s[/tex]
Step 5: Calculate the speed of the eagle after catching the pigeon.
Speed = [tex]$\sqrt{v_{f,x}^2 + v_{f,y}^2}$[/tex]
Speed = [tex]$\sqrt{15.2^2 + (-11.3)^2} m/s[/tex]
Speed = [tex]$19.0 m/s[/tex]
A monochromatic laser is exciting hydrogen atoms from the n=2 state to the n=5 state.
(A) What is the wavelength λ of the laser? Express your answer to three significant digits in nanometers.
(B) Eventually, all of the excited hydrogen atoms will emit photons until they fall back to the ground state. How many different wavelengths can be observed in this process?
(C) What is the longest wavelength λ_max that is observed? Express your answer to three significant digits in nanometers.
(D) What is the shortest wavelength λ_min observed? Express your answer to three significant digits in nanometers.
Answer:
a) λ = 435 nm , c) c) λ = 4052 nm, d) λ= 95 nm
Explanation:
A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.
En = -13,606 / n² [eV]
therefore the energy of the transition is
ΔE = E₅ -E₂
ΔE = 13.606 (1 / n₂² - 1 / n₅²)
ΔE = 13.606 (1/2² - 1/5²)
ΔE = 2,85726 eV
now let's use Planck's equation
E = h f
the speed of light is related to wavelength and frequencies
c = λ f
f = c /λ
E = h c /λ
λ = h c / E
let's reduce the energy to the SI system
E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J
let's calculate
λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹
λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition
initial state n = 5
final state n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ = h c / E
λ = 4052 nm
n = 5
final ΔE (eV) λ (nm)
level
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6
c) λ = 4052 nm
d) λ= 95 nm
The monochromatic laser has a wavelength of 434 nm. When excited hydrogen atoms fall back to the ground state, 10 different wavelengths can be observed ranging from 97.3 nm to 121.6 nm.
Explanation:A monochromatic laser is used to excite hydrogen atoms from the n=2 state to the n=5 state. The wavelength of the laser can be calculated using the Rydberg formula for hydrogen, 1 / λ = RH (1/n1^2 - 1/n2^2), where RH is the Rydberg constant for hydrogen (1.097×10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level. After calculating, we get λ = 434 nm.
Eventually, as excited hydrogen atoms fall back to the ground state, there are 10 different transitions possible, corresponding to 10 different wavelengths of light.
The longest wavelength λ_max is observed when the electron falls from n=2 to n=1 state, λ_max = 121.6 nm.
The shortest wavelength λ_min is observed when an electron falls from n=5 to n=1, λ_min = 97.3 nm.
Learn more about Excitation and Emission of Light in a Hydrogen Atom here:https://brainly.com/question/29056939
#SPJ3
A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume
Answer:
A) 100
B) 10
C) 0.01
Explanation:
Let
Fs = force on slave cylinder
Fm = force on master cylinder
Given that the designed to exert a force 100 times as large as the one put into it. That is
Fs = 100 Fm
Using Pascal's law
Fs/As = Fm/Am
100Fm/As = Fm/Am
100/As = 1/Am
Cross multiply
As= 100Am
As/Am = 100
b) What must be the ratio of their diameters
Using their areas ratio, area of a cylinder is πr^2
A = πr^2 = π(D/2)^2
π(Ds/2)^2 ÷ π(Dm/2)^2 = 100
(Ds/2)^2 ÷ (Dm/2)^2 = 100
(Ds/Dm)^2 = 100
Ds/Dm = 10
(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction
Let us consider the workdone at the input and at the output.
Work done = force × distance
Fs × Hs = Fm × Hm
100Fm × Hs = Fm × Hm
100Hs = Hm
100Hs/Hm = 1
Hs/Hm = 1/100
Hs/Hm = 0.01
An insulated piston–cylinder device contains 0.05 m3 of saturated refrigerant- 134a vapor at 0.8-MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 0.4 MPa. Determine (a) the final temperature in the cylinder and (b) the work done by the refrigerant
The final temperature of the refrigerant in the piston-cylinder device can be determined via lookup and interpolation from standard tables, based on the final pressure of 0.4 MPa. The work done by the refrigerant as it expands and drops in pressure can be calculated via the first law of thermodynamics, specifically accounting for changes in internal energy as work done by the system.
Explanation:This is an example of a reversible adiabatic expansion process involving a refrigerant (refrigerant-134a) in a piston-cylinder device. The process follows the behavior of ideal gases, and so we can make use of the first law of thermodynamics and other gas laws to determine the final temperature and the work done by the refrigerant.
To find the final temperature, we incorporate the specifics of the refrigerant-134a, which is defined by specific tables found in thermodynamics textbooks, providing details of temperature at various pressures. Interpolation between table values may be necessary to find the exact temperature at 0.4 MPa. Generally, the temperature will fall as the refrigerant expands and the pressure lowers, as defined by the ideal gas law (pV=nRT).
For the work done by the refrigerant, we'd need to apply the first law of thermodynamics, which in this case can be simplified as ΔU = W because there's no heat transfer in an adiabatic process (dQ=0). Changes in the internal energy of the gas would translate directly into work done on the piston. We calculate work using the formula -P(ΔV), where the negative sign indicates work being done by the system.
Learn more about Reversible adiabatic expansion here:https://brainly.com/question/33381819
#SPJ11
(a) The final temperature in the cylinder is approximately 18.0°C.
(b) The work done by the refrigerant during the expansion is approximately 7.72 kJ.
To solve this problem, we need to determine two things: the final temperature of the refrigerant-134a vapor and the work done by the refrigerant during the expansion.
Given that the process is reversible and the system is insulated, this suggests that the expansion process is isothermal for an ideal gas, but since refrigerant-134a is not an ideal gas, we will use the refrigerant tables to find the properties.
Given Data:
Initial volume [tex]\( V_1 \)[/tex] : 0.05 m³Initial pressure [tex]\( P_1 \)[/tex] : 0.8 MPaFinal pressure [tex]\( P_2 \)[/tex] : 0.4 MPaThe process is reversibleStep-by-Step Solution:
(a) Determine the Final Temperature in the Cylinder:
1. Find the Initial and Final States Using Refrigerant-134a Tables:
Initial State [tex](P1 = 0.8 MPa)[/tex]:Look up the refrigerant-134a property tables for [tex]\( P = 0.8 \)[/tex] MPa.
At [tex]\( P = 0.8 \)[/tex] MPa, the refrigerant-134a is in the saturated vapor region. The saturated temperature and properties at this pressure are:
Saturated temperature [tex]\( T_1 \): \( T_{\text{sat}}(0.8 \text{ MPa}) \approx 26.7^\circ \text{C} \)[/tex]Specific volume [tex]\( v_g \): \( v_{\text{g}}(0.8 \text{ MPa}) \approx 0.0769 \text{ m}^3/\text{kg} \)[/tex]Final State [tex](P2 = 0.4 MPa)[/tex] :Similarly, look up the refrigerant-134a property tables for [tex]\( P = 0.4 \)[/tex] MPa.
At [tex]\( P = 0.4 \)[/tex] MPa, the refrigerant-134a is also in the saturated vapor region. The properties at this pressure are:
Saturated temperature [tex]\( T_2 \)[/tex] : [tex]\( T_{\text{sat}}(0.4 \text{ MPa}) \approx 18.0^\circ \text{C} \)[/tex]Specific volume [tex]\( v_g \)[/tex] : [tex]\( v_{\text{g}}(0.4 \text{ MPa}) \approx 0.0917 \text{ m}^3/\text{kg} \)[/tex]The final temperature of the refrigerant-134a after the expansion, which is at 0.4 MPa, is approximately [tex]\( 18.0^\circ \text{C} \)[/tex].
(b) Determine the Work Done by the Refrigerant:
1. Calculate the Initial and Final Mass:
Using the specific volume [tex]\( v_g \)[/tex] to find the mass:
[tex]\[ m = \frac{V_1}{v_g (P_1)} \][/tex]
For [tex]\( P_1 = 0.8 \)[/tex] MPa:
[tex]\[ m = \frac{0.05 \text{ m}^3}{0.0769 \text{ m}^3/\text{kg}} \approx 0.650 \text{ kg} \][/tex]
For [tex]\( P_2 = 0.4 \)[/tex] MPa, verify if the mass is the same (which it is for an ideal gas in a closed system):
[tex]\[ V_2 = m \times v_g (P_2) = 0.650 \text{ kg} \times 0.0917 \text{ m}^3/\text{kg} \approx 0.0597 \text{ m}^3 \][/tex]
The new volume [tex]\( V_2 \)[/tex] can be verified as:
[tex]\[ V_2 = \text{initial volume} \times \frac{v_g (P_2)}{v_g (P_1)} \][/tex]
Which confirms our calculation.
2. Calculate the Work Done During Expansion:
For a reversible expansion in an insulated system (isoenthalpic process), the work done \( W \) can be approximated using:
[tex]\[ W = \int_{V_1}^{V_2} P \, dV \][/tex]
The work done for an ideal gas (but approximate here for refrigerants) can be simplified using:
[tex]\[ W = \text{P}_1 \times V_1 \ln \left( \frac{V_2}{V_1} \right) \][/tex]
For refrigerants, use specific volume:
[tex]\[ W = \text{P}_1 \times V_1 \left[ \frac{v_g (P_2)}{v_g (P_1)} - 1 \right] \][/tex]
Substituting:
[tex]\[ W = 0.8 \text{ MPa} \times 0.05 \text{ m}^3 \left( \frac{0.0917}{0.0769} - 1 \right) \][/tex]
[tex]\[ W = 0.8 \times 10^3 \text{ Pa} \times 0.05 \text{ m}^3 \left( 1.193 - 1 \right) \][/tex]
[tex]\[ W = 40 \text{ kJ} \times 0.193 \approx 7.72 \text{ kJ} \][/tex]
Aspartic acid is a polypeptide side chain found in proteins. The pKa of aspartic acid is 3.86. If this polypeptide were in an aqueous solution with a pH of 7, the side chain would have what charge? a) negative b) positive c) neutral d) there is no way to know
Answer: a) Negative
Explanation: According to the question: The pKa of aspartic acid is 3.86.
For acidic amino acids,
pH > pKa
When side chains are negatively charged, amino acid will be acidic
pH < pKa
That is, when side chains are uncharged, amino acid will be neutral
For aspartic acid, when this polypeptide were in an aqueous solution with a pH of 7
pH (7.0) > pKa (3.86)
Therefore the chains are negatively charged
Final answer:
In an aqueous solution at pH 7, the aspartic acid side chain would have a (a)negative charge because the pH of the environment is higher than the pKa of aspartic acid (3.86).
Explanation:
The question relates to the charge of the aspartic acid side chain in an aqueous solution at pH 7, given that the pKa of aspartic acid is 3.86. The pKa value represents the pH at which half of the aspartic acid side chains are deprotonated (negative charge) and half remain protonated (neutral). When the pH of the environment is higher than the pKa (as is the case here, pH 7 > pKa 3.86), the majority of the aspartic acid side chains will be deprotonated, thus carrying a negative charge. Consequently, in an aqueous solution at pH 7, the aspartic acid side chain would have (a)negative charge.
Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0.048 m. You wish the mass to have a kinetic energy of 14.0 Joules when released. How strong will the string need to be
Answer:
Tension in the string is equal to 58.33 N ( this will be the strength of the string )
Explanation:
We have given mass m = 1.7 kg
radius of the circle r = 0.48 m[tex]F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N[/tex]
Kinetic energy is given 14 J
Kinetic energy is equal to [tex]KE=\frac{1}{2}mv^2[/tex]
So [tex]\frac{1}{2}\times 1.7\times v^2=14[/tex]
[tex]v^2=16.47[/tex]
v = 4.05 m/sec
Centripetal force is equal to [tex]F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N[/tex]
So tension in the string will be equal to 58.33 N ( this will be the strength of the string )
Final answer:
The problem requires finding the tension in the string by using the given kinetic energy to calculate the velocity and then applying the centripetal force formula to find the required string strength.
Explanation:
The student question involves calculating the tension in a string when a mass is swung to obtain a certain kinetic energy. Considering the mass M has a kinetic energy of 14.0 Joules, we can use the relation for kinetic energy (KE) of a rotating body, which is KE = 1/2 m v^2, where 'm' is the mass and 'v' is the tangential velocity. From the given kinetic energy and mass, we can solve for v.
Once we have the velocity, we can use centripetal force formula, which is F = m v^2 / r, where 'm' is the mass, 'v' is the velocity, and 'r' is the radius. This formula will allow us to find the required strength of the string, which, in physics, is essentially the tension that the string must be able to withstand without breaking.
Note that the tension in the string will be equal to the centripetal force at the point of release when the mass is in horizontal circular motion.
A car goes from 100 m/s to a full stop for in 9.5 seconds. What is the acceleration?
Answer:Calculate displacement of an object that is not acceleration, given initial position and velocity.
Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.
Four men racing up a river in their kayaks.
Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr).
We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
Notation: t, x, v, a
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is
Δ
t
=
t
f
−
t
0
, taking
t
0
=
0
means that
Δ
t
=
t
f
, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,
x
0
is the initial position and
v
0
is the initial velocity. We put no subscripts on the final values. That is,
t
is the final time,
x
is the final position, and
v
is the final velocity. This gives a simpler expression for elapsed time—now,
Δ
t
=
t
. It also simplifies the expression for displacement, which is now
Δ
x
=
x
−
x
0
. Also, it simplifies the expression for change in velocity, which is now
Δ
v
=
v
−
v
0
. To summarize, using the simplified notation, with the initial time taken to be zero,
Explanation:
A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit specific enthalpy of 297.31 kj/kg. For negligible heat transfer with the surroundings, the exit velocity is
a) 223 m/s
b) 197 m/x
c) 90 m/s
d) 70 m/s
Answer: c) 90 m/s
Explanation:
Given
Invest velocity, v1 = 250 m/s
Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg
Outlet specific enthalpy, h2 = 297.31 kJ/kg = 297310 J/kg
Outlet velocity, v2 = ?
0 = Q(cv) - W(cv) + m[(h1 - h2) + 1/2(v1² - v2²) + g(z1 - z2)]
0 = Q(cv) + m[(h1 - h2) + 1/2(v1² - v2²)]
0 = [(h1 - h2) + 1/2(v1² - v2²)]
Substituting the values of the above, we get
0 = [(270110 - 297310) + 1/2 ( 250² - v²)
0 = [-27200 + 1/2 (62500 - v²)]
27200 = 1/2 (62500 - v²)
54400 = 62500 - v²
v² = 62500 - 54400
v² = 8100
v = √8100
v = 90 m/s
A proton moves through a magnetic field at 26.1 % of the speed of light. At a location where the field has a magnitude of 0.00667 T and the proton's velocity makes an angle of 139 ∘ with the field, what is the magnitude F B of the magnetic force acting on the proton? Use c = 2.998 × 10 8 m/s for the speed of light and e = 1.602 × 10 − 19 C as the elementary charge.
Answer:
[tex]5.48\cdot 10^{-14} N[/tex]
Explanation:
When a charged particle is moving in a region with a magnetic field, the particle experiences a force perpendicular to its direction of motion. The magnitude of this force is given by
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is its velocity
B is the strength of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
In this problem we have:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]v=0.261 c[/tex] is the speed of the proton, where
[tex]c=2.998\cdot 10^8 m/s[/tex] is the speed of light
[tex]B=0.00667 T[/tex] is the strength of the magnetic field
[tex]\theta=139^{\circ}[/tex] is the angle between the direction of the proton and the magnetic field
Substituting, we find the magnitude of the force:
[tex]F=(1.6\cdot 10^{-19})(0.261\cdot 2.998\cdot 10^8)(0.00667)(sin 139^{\circ})=5.48\cdot 10^{-14} N[/tex]
The magnitude of the magnetic force acting on the proton is [tex]5.46*10^{-14} N[/tex]
Force on charge :
When a charge particle is moving in magnetic field.
Then force is given as, [tex]F=qvBsin(\theta)[/tex]
where
[tex]q[/tex] is the charge of the particle[tex]v[/tex] is velocity[tex]B[/tex] is the strength of the magnetic field[tex]\theta[/tex] is the angle between the direction of [tex]v[/tex] and [tex]B[/tex]Given that, [tex]q=1.6*10^{-19}C,v=0.261*3*10^{8}=7.8*10^{7} ,B=0.00667T,\theta=139[/tex]
Substitute all values in above relation.
[tex]F=1.6*10^{-19}*7.8*10^{7} *0.00667T*sin(139)\\\\F=5.46*10^{-14} N[/tex]
Learn more about the magnetic force here:
https://brainly.com/question/2279150
What is the magnitude of the rate of change of the magnetic field inside the coil? You will need to calculate this algebraically before you can get a number. Write an expression for the magnetic field due to the wire at the location of the coil. Use the approximate formula, since the wire is very long. Remember the chain rule, and remember that ⅆ ⅆ .
Answer:
The answer to this questions are, (a) 0.685 * 10^-4 T/sec (b) 0.946 * 10^-6V
Explanation:
Solution
Recall
radius r = 0.02m
N = 11 turns
Instant I = 3 amperes
Velocity v =3.3m/s
x = 0.17 m
(a) What is the magnitude of the rate of change of the magnetic field inside the coil
db/dt =μ₀T/2π x²
Thus,
= 4π * 10^-7* 3* 3.3/2π * 0.17²
=685.12* 10^-7
which is now,
0.685 * 10^-4 T/sec
(b) What is the magnitude of the voltmeter reading.
μ = N (db/dt)πr²
Note: this includes all 11 turns of the coil
Thus,
= 11 * 0.6585* 10^-4* 3.14 * (0.02)²
= 946 * 10^-7
which is = 0.946 * 10^-6V
Note: Kindly find an attached copy or document of the complete question of this exercise
The rate of change of a magnetic field in a coil can be determined using Faraday's law and the chain rule. Starting with the formula for the magnetic field created by current in a straight wire, we differentiate with respect to time to get the rate of change.
Explanation:The magnitude of the rate of change of the magnetic field inside a coil can be represented algebraically using Faraday's law of induction. Starting with the equation B = μ * I / 2λr (where I is the current, r is the distance to the wire, and μ is the permeability of free space), we can find the rate of change of the magnetic field using the chain rule.
Faraday's law is represented as E = - dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is time. Since the magnetic flux Φ is the product of the magnetic field (B) and the area enclosed by the loop (A), we can express this as d/dt (B * A). By applying the chain rule, we can find the rate of change of the magnetic field in the coil.
Last, magnetic field B is determined by Ampère's law, often used in calculations involving magnetic fields around conductors and coils. For a long straight wire, field lines form concentric circles around the wire, following the right-hand rule.
Learn more about Rate of change of magnetic field here:https://brainly.com/question/40242415
#SPJ3
how is velocity different from speed
Answer: velocity has a direction, while speed is the distance travelled by an object.
Explanation:
Answer:
Velocity includes Displacement instead of distance.
Explanation:
Velocity's formula is change in position divided(displacement) by change in time(Delta x/Delta t). Whereas speed is distance over change in time(d/Delta t).
A spaceship is traveling toward the earth from the space colony on Asteroid 1040A. The ship is at the halfway point of the trip, passing Mars at a speed of 0.9c relative to the Mars frame of reference. At the same instant, a passenger on the spaceship receives a radio message from her boyfriend on 1040A and another from her sister on earth. According to the passenger on the ship, were these messages sent simultaneously or at different times? If at different times, which one was sent first? Explain your reason
Answer:
The message (signal) coming from the earth was sent first.
Explanation:
The radio messages reach Mars simultaneously, because the distance from Earth to Mars and that of from Mars to Asteroid is same. But the passenger in the spaceship is moving relative to Mars towards Earth and hence, the message from Earth reaches first.
According to passenger frame, the message (signal) coming from the Earth was sent first compared to message coming from Asteroid.
The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz.
a. What is the spring constant of the spring on which the head is mounted?
b. suppose the head is pushed 2.0 cm against the spring, then released. What is the head's maximum speed as it oscillates?
c. the amplitude of the head's oscillations decreases to 0.5cm in 4.0s. What is the head's damping constant?
Answer:
(a) 9.375 N/m
(b) 0.5024 m/s
(c) 0.01 kg/s
Explanation:
mass of head, m = 15 g = 0.015 kg
frequency, f = 4 Hz
Time period, T = 1 / f = 0.25 s
Let k is the spring constant.
(a)
The formula for the time period is
[tex]T=2\pi\sqrt{\frac{m}{K}}[/tex]
[tex]0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}[/tex]
[tex]0.04=\sqrt{\frac{0.015}{K}}[/tex]
K = 9.375 N/m
(b)
Amplitude, A = 2 cm
Let ω is the angular velocity.
Maximum velocity, v = A ω = A x 2πf
v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s
(c)
Let b is the damping constant.
A(t = 4s) = 0.5 cm
Ao = 2 cm
Using the formula of damping
[tex]\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}[/tex]
[tex]\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}[/tex]
[tex]0.25=e^{-133.3 b}[/tex]
Taking natural log on both the sides
ln (0.25) = - 133.3 b
- 133.3 b = - 1.386
b = 0.01 kg/s
This question involves the concepts of simple harmonic motion, spring constant, and amplitude.
a) The spring constant of the spring is "9.47 N/m".
b) The maximum speed of the head is "0.5 m/s".
c) The damping constant is "0.01 kg/s".
a)
We can find the spring constant of the spring by using the formula of frequency in the simple harmonic motion:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where,
f = frequency = 4 Hz
k = spring constant = ?
m = mass = 15 g = 0.015 kg
Therefore,
[tex]4\ Hz=\frac{1}{2\pi}\sqrt{\frac{k}{0.015\ kg}}\\\\(16\ Hz^2)(4\pi^2)(0.015\ kg)=k\\[/tex]
k = 9.47 N/m
b)
Maximum speed is simply given by the following formula:
[tex]v=A\omega[/tex]
where,
v = maximum speed = ?
A = Amplitude = 2 cm = 0.02 m
ω = angular freuency = 2πf
Therefore,
[tex]v=A(2\pi f)=(0.02\ m)(2\pi)(4\ Hz)[/tex]
v = 0.5 m/s
c)
using the following equation to find out the damping constant:
[tex]ln(\frac{A}{A_o})=-\frac{bt}{2m}[/tex]
where,
A = amplitude at t = 4 s = 0.5 cm
A₀ = initial amplitude = 2 cm
b = damping constant = ?
t = time = 4 s
m = mass = 15 g = 0.015 kg
Therefore,
[tex]ln(\frac{0.5\ cm}{2\ cm})=-\frac{b(4\ s)}{2*0.015\ kg}[/tex]
[tex]\frac{(-1.386)(2)(0.015\ kg)}{4\ s}=-b[/tex]
b = 0.01 kg/s
Learn more about simple harmonic motion here:
https://brainly.com/question/17315536?referrer=searchResults
Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual image of the mite. If the focal distance of the objective lens is shortened, then its magnification decreases. The converging objective lens forms a virtual image of the mite. If the focal distance of the objective lens is lengthened, then its magnification increases. The image formed by the objective lens should be located just inside the focal point of the eyepiece lens
Answer:
a) True. The image of the mite is virtual
e) True. The image must be within the focal length of the eyepiece len
Explanation:
Let's review the general characteristics of compound microscopes
Formed by two converging lenses
Magnification is
M = -L/fo 0.25/fe
Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length
Let's review the claims
a) True. The image of the mite is virtual
b) False. The effect is the opposite of the magnification equation
c) False. The objective lens forms a real image
d) False. As the seal distance increases the magnification decreases
e) True. The image must be within the focal length of the eyepiece len
A compound microscope uses an objective lens to create a real, inverted image larger than the object, and an eyepiece (ocular) to magnify this image further into a virtual image. The correct statements include that the eyepiece forms a virtual image, and the image from the objective lens should be within the eyepiece's focal length for further magnification.
Explanation:To understand how a compound microscope works, we analyze the function of its two lenses: the objective lens and the eyepiece. Initially, an object just beyond the objective lens's focal length (fobj) creates a real, inverted image that is larger than the object. This initial image acts as the object for the eyepiece lens, or ocular, which is placed within its own focal length (feye) to further magnify the image, effectively functioning as a magnifying glass. The resulting image is a virtual, larger, and further magnified version of the initial image, making it more comfortable for viewing as the eye is relaxed when looking at distant objects.
If we consider the statements provided, the correct statements are as follows:
https://brainly.com/question/35060850
#SPJ3
A 20 kg box on a horizontal frictionless surface is moving to the right at a speed of 4.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then oscillates back and forth. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero. ____ Positive ____ Negative ____ Zero
Answer:
Work done by the spring is negative
Explanation:
We can answer this question by thinking what is the force acting on the box.
In fact, the force acting on the box is the restoring force of the spring, which is given by Hooke's Law:
[tex]F=-kx[/tex]
where
k is the spring constant
x is the displacement of the box with respect to the equilibrium position of the spring
The negative sign in the equation indicates that the direction of the force is always opposite to the direction of the displacement: so, whether the spring is compressed or stretched, the force applied by the spring on the box is towards the equilibrium position.
The work done by the restoring force is also given by
[tex]W=Fx cos \theta[/tex]
where
F is the restoring force
x is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and the displacement
Here we know that the force is always opposite to the displacement, so
[tex]\theta=180^{\circ} \rightarrow cos \theta =-1[/tex]
Which means that the work done by the spring is always negative, since the direction of the restoring force is always opposite to the direction of motion.
The work done by the spring on the box as it compresses is negative, due to the force applied by the spring being in the opposite direction of the box's displacement.
Explanation:For the given problem, the work done by the spring on the box is negative. This is because the spring is exerting a force in the opposite direction to that of the box's motion, leading to a decrease in the kinetic energy of the box. According to the work-energy principle, if the force applied by the spring opposes the direction of motion, the work done is negative. When the spring compresses, it stores elastic potential energy, which is a conversion of the box's kinetic energy. The spring force acts to the left (assuming the right is the positive direction), while the displacement of the box due to that force is to the right, resulting in negative work done by the spring.
It’s mid-afternoon on a lovely, sunny day; it’s 60°F, and the sun is 60° above the horizon. Rays from the sun strike the still surface of a shallow pond that is 20 cm deep. The rays of sun cast a shadow of a stick that is completely submerged, and stuck in the sandy bottom of the pond. If 10 cm of the stick is above the sand, what is the length of the shadow?
Answer:
The correct answer is 5.77 cm
Explanation:
Solution
From the question given, let us recall the following,
A sunny day = 60°F and
The sun above the horizon is = 60°
The surface of a shallow pond is = 20 cm deep
The stick above sand = 10 cm
The next step is to find the length of the shadow
Now,
let CD = stick
BC= Length of shadow
Thus,
CD/BC = Tan 60°
BC = CD/Tan 60°
= 10/√3 = 5.77 cm
Therefore, the length of the shadow is 5.77 cm
It's a hot and sunny day and you decide to go for a drive. When you open your car door, it's a lot hotter than outside. Why is that
A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude a. What is the tension in the cable between the drum and the bucket
Final answer:
The tension in the cable connecting the drum to the bucket is equal to the mass of the bucket times the difference between the acceleration due to gravity and the bucket's downward acceleration.
Explanation:
To determine the tension in the cable between the drum and the bucket, we can apply Newton's second law to the hanging mass. The net force acting on the bucket is the difference between the weight of the bucket and the tension in the rope:
[tex]F_{net}[/tex] = mg - T
Since we know the bucket has a downward acceleration a, we can write Newton's second law as:
ma = mg - T
Where m is the mass of the bucket, g is the acceleration due to gravity, and T is the tension in the rope. We can rearrange the equation to solve for T:
T = mg - ma
T = m(g - a)
So the tension in the cable is equal to the mass of the bucket times the difference between the acceleration due to gravity and the downward acceleration of the bucket.
A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, away. The sun emits a power of 3.8 * 1026 W. Using this information, at what distance would a candle that emits a power of 0.20 W just be visible?
To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The distance at which the candle would be visible is approximately 2.1 million kilometers.
Explanation:To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The luminosity of the Sun is 3.8 * 10^26 W. The candle's luminosity can be calculated using the ratio of its power to the power of the Sun. Luminosity is inversely proportional to the distance squared, so we can set up an equation with the ratio of the candle's luminosity to the Sun's luminosity equal to the ratio of the distance at which the candle is visible to the distance at which the Sun is barely visible:
(0.20 W) / (3.8 * 10^26 W) = (6.6 * 10^16 m)^2 / (x)^2
Cross-multiplying and solving for x, we find that the distance at which the candle would be visible is approximately 2.1 * 10^9 m, or 2.1 million kilometers.
Learn more about distance at which the candle would be visible here:https://brainly.com/question/21730331
#SPJ12
Approximately 4785 meters is the distance at which a candle would just be visible.
To solve this problem, we need to use the inverse square law for the intensity of light. The apparent brightness of a light source decreases with the square of the distance from the observer. The formula relating the brightness (intensity I) at a distance d is given by:
[tex]\[ I = \frac{P}{4 \pi d^2} \][/tex]
First, we find the intensity I at the distance where the sun-like star is barely visible:
[tex]\[ I = \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} \][/tex]
Plugging in the values:
[tex]\[ I = \frac{3.8 \times 10^{26}}{4 \pi (6.6 \times 10^{16})^2} \][/tex]
Now we need to find the distance [tex]\( d_{\text{candle}} \)[/tex] at which the candle with power [tex]\( P_{\text{candle}} \)[/tex] would have the same intensity I:
[tex]\[ I = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]
Setting the intensities equal:
[tex]\[ \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]
Solving for [tex]\( d_{\text{candle}} \)[/tex]:
[tex]\[ \frac{P_{\text{star}}}{d_{\text{star}}^2} = \frac{P_{\text{candle}}}{d_{\text{candle}}^2} \][/tex]
[tex]\[ d_{\text{candle}}^2 = d_{\text{star}}^2 \frac{P_{\text{candle}}}{P_{\text{star}}} \][/tex]
[tex]\[ d_{\text{candle}} = d_{\text{star}} \sqrt{\frac{P_{\text{candle}}}{P_{\text{star}}}} \][/tex]
Plugging in the values:
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{2 \times 10^{-1}}{3.8 \times 10^{26}}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{5.26 \times 10^{-28}} \][/tex]
[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \times 7.25 \times 10^{-14} \][/tex]
[tex]\[ d_{\text{candle}} = 4.785 \times 10^3 \][/tex]
[tex]\[ d_{\text{candle}} \approx 4785 \, \text{meters} \][/tex]
Thus, a candle that emits a power of 0.20 W would just be visible at a distance of approximately 4785 meters.
How would you classify the wave?
Answer:
One way to categorize waves is on the basis of the direction of movement of the individual particles of the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three notable categories: transverse waves, longitudinal waves, and surface waves.
I've read about it but never seen one. The way I understand it, it's a coordinated physical motion executed by a great number of people, as in a large crowd at a sporting event, timed so that it appears to propagate from one end of the crowd to the opposite end.
I would classify it as a cooperative community activity, involving liberty, equality, and fraternity, executed for the common good.
Calculate the magnitude of the magnetic force on a 270 m length of wire stretched between two towers and carrying a 150 A current. The Earth's magnetic field of 4.0×10−5 T makes an angle of 80 ∘ with the wire
Answer:
Force on the wire is equal to 15.95 N
Explanation:
We have given length of the wire l = 270 m
Current flowing in the wire i = 150 A
Magnetic field [tex]B=4\times 10^{-5}T[/tex]
Angle between magnetic field and wire [tex]\Theta =80^{\circ}[/tex]
We have to find the force on the wire
Force on current carrying conductor is equal to [tex]F=IBlsin\Theta[/tex]
So [tex]F=150\times 4\times 10^{-5}\times 270\times sin80^{\circ}=15.95N[/tex]
So force on the wire is equal to 15.95 N
Using the formula F = IlB sin(θ), with I = 150 A, l = 270 m, B = 4.0×10⁻⁵ T, and θ = 80°, the magnitude of the magnetic force on the wire is calculated to be approximately 158.76 Newtons.
Explanation:To calculate the magnitude of the magnetic force on the wire, we use the formula F = IlB sin(θ), where F represents the force, I is the current, l is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field. In this scenario, I = 150 A, l = 270 m, B = 4.0×10⁻⁵ T, and θ = 80°.
First, we calculate the sine of the angle:
sin(80°) ≈ 0.9848Plugging the values into the force equation gives:
F = (150 A) × (270 m) × (4.0×10⁻⁵ T) × 0.9848
F = 158.76 N (rounded to two decimal places)
Hence, the magnitude of the magnetic force on the wire is approximately 158.76 Newtons.
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 5.20 m and, in starting, rotates according to θ = 0.170t2, where t is in seconds and θ is in radians. When t = 4.70 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?
Answer:
A) angular velocity; ω = 1.598 rad/s
B) linear velocity;V = 8.31 m/s
C) Tangential Acceleration;a_t = 1.768 m/s²
D) Radial Acceleration;a_r = 13.28 m/s²
Explanation:
We are given that;
Radius; r = 5.2m
Time;t = 4.7 sec
θ = 0.170t²
Thus, angular acceleration would be the second derivative of θ which is d²θ/dt²
Thus,α = d²θ/dt² = 0.34 rad/s²
A) Formula for angular velocity is;
ω = αt
Where α is angular acceleration and t is time.
Thus;ω = 0.34 x 4.7
ω = 1.598 rad/s
b) formula for linear velocity is given by; V = ωr
We have ω = 1.598 rad/s and r = 5.2m
Thus; V = 1.598 x 5.2
V = 8.31 m/s
c) formula for tangential acceleration is;
a_t = αr
a_t = 0.34 x 5.2
a_t = 1.768 m/s²
d) formula for radial acceleration is;
a_r = rω²
a_r = 5.2 x 1.598²
a_r = 13.28 m/s²
To find the magnitudes of the astronaut's angular velocity, linear velocity, tangential acceleration, and radial acceleration, differentiate the angular position equation, multiply the radius and the angular velocity to find the linear velocity, and use the rate of change of linear velocity to find the tangential acceleration. The radial acceleration is the product of the angular velocity and the linear velocity.
Explanation:The angular velocity can be found by differentiating the angular position equation with respect to time. The linear velocity is equal to the product of the radius and the angular velocity. The tangential acceleration is equal to the rate of change of linear velocity, while the radial acceleration is equal to the product of the angular velocity and the linear velocity.
Learn more about angular velocity here:https://brainly.com/question/30733452
#SPJ11