Answer:
Explanation:
moment of inertia of the door M = 1 /3 m ( l² + b² + d² )
= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )
= 1 / 3 x 7.5 x 6.2909
I = 15.7272.
moment of inertia of the door + small toy
15.7272 + .12 x .5²
= 15.7272 + .03
= 15.7572
Applying conservation of angular momentum law
mvr = M ω
m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.
.12 x v x .5 = 15.7572 x .19
v = 49.89 m /s .
As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a channel in a microprocessor. If we know that an electron is somewhere along the 50 nm length of the channel, what is ∆vx? If we treat the elec- tron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it take to traverse the channel?
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
To solve this problem you use the Heisenberg's uncertainty principle, that is given by:
[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]
where h is the Planck's constant (6.62*10^-34 J s).
If you assume that the mass of the electron is constant you have:
[tex]\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}[/tex]
you use the value of the mass of an electron (9.61*10^-31 kg), and the uncertainty in the position of the electron (50nm), in order to calculate ∆x∆v and ∆v:
[tex]\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s[/tex]
[tex]\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}[/tex]
If the electron is a classical particle, the time it takes to traverse the channel is (by using the edge of the uncertainty in the velocity):
[tex]t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s[/tex]
Un ciclista recorre una pista recta de ida y vuelta, en este recorrido:
A) La distancia total recorrida es cero.
B) La aceleración es distinta de cero.
C) El tiempo total es cero.
D) La rapidez es cero.
E) El desplazamiento total es cero.
Answer:
a) falso
b) verdadero
c) falso
d) falso
e) verdadero
Explanation:
A) FALSO: la distancia recorrida es independiente del punto de partida y del punto de llegada
B) VERDADERO: para que el ciclista recorra la pista de ida y vuelta es necesario que su velocidad cambie, por lo tanto la aceleración es diferente de cero.
C) FALSO
D) FALSO: en todo momento hay una distancia recorrida en un tiempo en específico, es decir, una rapidez.
E) VERDADERO: el despazamiento sí depende de la distancia entre el punto de llegada y el punto de salida.
What is one way to increase the amplitude if a wave in a medium?
A. By applying a vibration at the natural frequency of the medium
B. By increasing the velocity of the wave source
C. By moving the source of the wave away from the receptor or observer
D. By increasing the frequency of the wave source
Answer: c
Explanation:
How does a thermos keep hot drinks hot?
A: By absorbing radiant energy that might enter the thermos
B: By limiting the amount of thermal energy that flows out of the drink
C: By limiting the amount of cold that flows into the drink
D: By transmitting radiant energy into the thermos
Answer:
the answer is B
Explanation:
hope this helps
Answer:
b
Explanation:
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 30.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 99.0 kg, perfectly absorbing block.
What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
4. An electric toaster is rated at 500 watts. Determine the amount of electrical
energy it converts to heat energy in one minute.
80.3 J
B)
500 J
C)
3,000 J
D)
30,000 J
Answer:60000 joules
Explanation:
power=500watts
Time=2minutes=120seconds
Heat energy=power x time
Heat energy=500 x 120
Heat energy=60000J
An electric toaster rated at 500 watts converts electrical energy into heat energy at a rate of 500 joules per second. To find the energy used in one minute, we simply multiply the wattage by the seconds in a minute, resulting in 30,000 joules.
Explanation:An electric toaster rated at 500 watts converts electrical energy into heat energy at a rate of 500 joules per second. Since watts are joules per second, to find out how much energy you would use in one minute, you need to multiply the number of watts by the number of seconds in one minute. There are 60 seconds in a minute so the computation is 500 joules/second × 60 seconds = 30000 joules. Therefore, in one minute, a 500-watt toaster will convert 30,000 joules of electrical energy into heat energy.
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You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work on water resources project. The department will be overseeing the construction of a dam to create a large fresh water lake that will be approximately 17 meters deep. A horizontal pipe 1.2 meters long and 3 cm in diameter will pass through the dam at a depth of 7 meters to allow for release of the water in emergencies and for sampling. In normal situations, a plug will secure the pipe opening. 1)What will be the total force on the left side of the plug
Answer:
Check the explanation
Explanation:
1) Pressure acting on the plug = Patm + P
Pressure = Patm + rho*g*h (Here h = D2)
Pressure = 101325 + 1000*9.8*7
Pressure = 169925 Pa
so, Force = PA
Force = 169925*pi*0.0152
Force = 120.1 N
Migratory birds are able to use the earth's magnetic field to navigate even when clouds and darkness prevent them from having visual references for their flight. The range of sensitivity of such birds extends to magnetic fields as small as about a third of the earth's natural field. If such a bird is flying past a power line which carries 105 amps, and if we assume that the minimum field detectable at 60 Hz is the same as the minimum field detected at DC, at what distance could the bird detect the presence of the power line? Why will the real answer be much smaller? (think about how power lines are arranged)
Answer:
The receptors that sense the Earth's magnetic field are probably located in the birds' eyes. Now, researchers at Lund University have studied different proteins in the eyes of zebra finches and discovered that one of them differs from the others: only the Cry4 protein maintains a constant level throughout the day and in different lighting conditions
Explanation:
Final answer:
The bird will be able to detect the power line at a much smaller distance than calculated due to the concentration of the magnetic field close to the wire.
Explanation:
The distance at which a migratory bird could detect the presence of a power line can be calculated using the equation for the magnetic field produced by a long straight wire. However, since the bird's sensitivity to magnetic fields is much smaller than the field produced by the power line, the real answer will be much smaller than the calculated distance. This is because power lines are typically arranged in such a way that the magnetic field they produce is concentrated close to the wire, diminishing rapidly with distance.
Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and three on the right (total mass mp=10.1 kgmp=10.1 kg ). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque ( ????pτp ) produced by the pile of rocks? (Enter a positive value.)
Answer:
Torque = 35.60 N.m (rounded off to 3 significant figures.
Explanation:
Given details:
The mass of the rock on the left, ms = 2.25 kg
The total mass of the rocks, mp = 10.1 kg
The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m
(a) The torque produced by the pile of rock, T = F*rp = m*g*rp
Torque = 9.8*0.360*10.1 = 35.6328
Torque = 35.60 N.m (rounded off to 3 significant figures).
A cyclist is moving up a slope that is at an angle of 19 to the horizontal. The mass of the cyclist and the bicycle is 85 kg. What is the component of the weight of the cyclist and bicycle parallel to the slope and what is the normal reaction force on the bicycle from the slope?
Answer:
Explanation:
Given
Slope of inclination [tex]\theta =19^{\circ}[/tex]
Mass of cyclist and bicycle is [tex]m=85\ kg[/tex]
When cyclist is going up then there is two components of its weight , one is parallel to inclination and other is perpendicular to inclination
Weight of person is mg
it can be resolved in [tex]mg\cos \theta[/tex] and [tex]mg\sin \theta [/tex]
[tex]mg\cos \theta[/tex] is perpendicular to the inclination
and [tex]mg\sin \theta[/tex] is parallel to inclination as shown in diagram
Parallel component [tex]=mg\sin \theta=85\times 9.8\times \sin 19=271.2\ N[/tex]
Normal reaction [tex]N=mg\cos \theta =85\times 9.8\times \cos 19=787.61\ N[/tex]
The normal reaction on the bicycle by the inclined slope is 787.62 N.
The parallel on the bicycle along the inclined slope is 271.2 N.
The given parameters;
angle of the slope, = 19mass of the bicycle, m = 85 kgThe normal reaction on the bicycle by the inclined slope is calculated as; follows;
[tex]F_n = W cos \theta\\\\F_n = mg \ cos \theta \\\\F_n = 85 \times 9.8 \times cos \ (19)\\\\F_n = 787.62 \ N[/tex]
The parallel on the bicycle along the inclined slope is calculated as
[tex]F_x = Wsin\theta\\\\F_x = mg \ sin \ \theta \\\\F_x = 85 \times 9.8 \times sin(19) \\\\F_ x = 271 .2 \ N[/tex]
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A cylindrical bar of metal having a diameter of 21.0 mm and a length of 210 mm is deformed elastically in tension with a force of 46800 N. Given that the elastic modulus and Poisson's ratio of the metal are 60.9 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number. (a) 0.443 mm (b) mm
Answer:
The the elongated length is [tex]\Delta L = 0.4 \ mm[/tex]
The change in diameter is [tex]\Delta d = - 0.0136\ mm[/tex]
Explanation:
From the question we are told that
The diameter of the cylindrical bar is [tex]d = 21.0 \ mm = \frac{21}{1000} = 0.021 \ m[/tex]
The length of the cylindrical bar is [tex]L= 210 \ mm = 0.21 \ m[/tex]
The force that deformed it is [tex]F = 46800 \ N[/tex]
Elastic modulus is [tex]E = 60.9 \ GPa = 60.9 *10^{9}Pa[/tex]
The Poisson's ratio is [tex]\mu = 0.34[/tex]
Generally elastic modulus is mathematically represented as
[tex]E = \frac{\sigma }{\epsilon}[/tex]
Where
[tex]\epsilon[/tex] is the strain which is mathematically represented as
[tex]\epsilon = \frac{L}{\Delta L}[/tex]
Where [tex]\Delta L[/tex] is the elongation length
[tex]\sigma[/tex] is the stress on the cylinder which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Where F is the force and
A is the area which is calculated as
[tex]A = \frac{\pi} {4} d^2[/tex]
Substituting values
[tex]A = \frac{\pi}{4} * (0.021)[/tex]
[tex]A = 0.000346 \ m^2[/tex]
So the stress is
[tex]\sigma = \frac{46800}{0.000346}[/tex]
[tex]\sigma = 1.35 *10^{8} \ N \cdot m^2[/tex]
Thus the elastic modulus is
[tex]E = \frac{1.35 *10 ^{8}}{\frac{\Delta L}{L} }[/tex]
making [tex]\Delta L[/tex] the subject
[tex]\Delta L = \frac{EL}{1.35 *10^{8}}[/tex]
Substituting values
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{1.35 *10^{8}}[/tex]
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{60.9*10^{9}}[/tex]
[tex]\Delta L = 0.0004 \ m[/tex]
Converting to mm
[tex]\Delta L = 0.0004 * 1000[/tex]
[tex]\Delta L = 0.4 \ mm[/tex]
Generally the poisson ratio is mathematically represented as
[tex]\mu = - \frac{\frac{\Delta d }{d} }{\frac{\Delta L }{L} }[/tex]
The negative sign indicate a decrease in diameter as a result of the force
making [tex]\Delta d[/tex] the subject
[tex]\Delta d = - \mu * \frac{\Delta L }{L } * d[/tex]
Substituting values
[tex]\Delta d = - 0.34 * \frac{0.0004 }{0.210 } * 0.021[/tex]
[tex]\Delta d = - 1.36 *10^{-5} \ m[/tex]
Converting to mm
[tex]\Delta d = - 0.0136\ mm[/tex]
2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initial temperature at 1336K. The solid and liquid can only exchange heat with each other. What kind of analysis do you need to perform in order to determine whether, once thermal equilibrium is reached, the mixture will be entirely solid or in a mixed solid/liquid phase?
Answer:
Explanation:
The specific heat of gold is 129 J/kgC
It's melting point is 1336 K
It's Heat of fusion is 63000 J/kg
Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,
The first is E1 = 63000 J/kg x 1.5 = 94500 J
the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J
Therefore, all solid is not correct. You will have a mixture of solid and liquid.
For more detail, the difference between E1 and E2 is 7812 J, and that will melt
7812/63000 = 0.124 kg of the solid gold
The question involves a thermodynamic analysis using heat transfer and phase change principles to determine if a mixture of solid and liquid gold will become entirely solid or remain mixed upon reaching thermal equilibrium.
Explanation:To determine whether the mixture of 2.0 kg of solid gold at 1000K and 1.5 kg of liquid gold at 1336K will be entirely solid or in a mixed solid/liquid phase upon reaching thermal equilibrium, a thermodynamic analysis involving the principles of heat transfer and phase change is required. This analysis will utilize the concept of conservation of energy, accounting for the specific heats of solid and liquid gold as well as the latent heat of fusion if a phase change occurs. The calculations will involve setting the heat lost by the liquid gold equal to the heat gained by the solid gold until both reach the same temperature. If the final temperature is below the melting point of gold without all the solid gold melting, the result will be a mixture. If all the solid melts before reaching the equilibrium temperature, the final state will be all liquid. It's critical to remember that the process must comply with the melting point of gold, which is 1064K.
5. A 30.0-g rifle bullet traveling 185 m/s embeds itself in a 3.15-kg pendulum hanging on a 2.85-m-long string, which makes the pendulum swing upward in an arc. Determine: a. The vertical component of the pendulum’s maximum displacement. (10pts) b. The horizontal component of the pendulum’s maximum displacement. (10pts) c. The angle of the pendulum’s maximum displacement with the vertical. (10pts)
Answer:
Explanation:
We shall apply law of conservation of momentum to calculate the common velocity of pendulum and bullet after collision
m v = ( m + M ) V
m is mass of bullet M is mass of pendulum , v is velocity of bullet and V is their final velocity after collision.
V = m v / ( m + M )
= .030 x 185 / 3.18
= 1.745 m /s
The kinetic energy of bullet+pendulum will be converted into potential energy.
1/2 (m+M) V² = ( m + M ) g h
h is height by which pendulum and bullet rises after collision
Putting the values
.5 x 3.18 x 1.745² = 3.18 x 9.8 x h
h = .155 m
If θ be the angle that the pendulum is making at deflected position
l - lcosθ = h , l is length of pendulum
2.85 ( 1 - cosθ) = .155
1 - cosθ = .054
cosθ = .946
θ = 19 degree
Horizontal displacement = 2.85 sin 19
= .9278 m
92.78 cm
Using the principle of conservation of momentum and kinetic energy transformation, we calculate the vertical displacement of the pendulum after the bullet embeds. This is then used to compute the horizontal displacement using the Pythagorean theorem. Finally, the angle of displacement is determined with the trigonometric principle of arcsine.
Explanation:To solve this question, we will use the principle of conservation of momentum to the bullet and pendulum system. After the bullet embeds into the pendulum, they move together, and due to the conservation of momentum, the joint velocity will be lower than the initial velocity of the bullet. We can calculate this joint velocity using the formula: (Mass of bullet x bullet velocity + Mass of pendulum x pendulum velocity) / (Mass of bullet + Mass of pendulum).
Next, we calculate the kinetic energy of the bullet-pendulum system after the bullet embeds itself using the formula: 1/2 x Mass of system x Velocity of system². This kinetic energy is then transformed into potential energy as the pendulum swings upward, which can be calculated using the formula: Mass of system x gravitational acceleration (9.81 m/s²) x vertical displacement.
We can solve for vertical displacement by equating the kinetic and potential energies. The horizontal displacement can then be calculated by using the Pythagorean theorem, where the hypotenuse is the length of the pendulum string, and one side is the length of the string less the vertical displacement (calculated earlier). Using these, the horizontal displacement is calculated as the square root of (length of string² - vertical displacement²).
The angle of maximum displacement with the vertical can then be calculated using the principle of trigonometry, namely the Arcsine formula, where arcsin (vertical displacement/length of string).
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A 17.0 m long, thin, uniform steel beam slides south at a speed of 28.0 m/s. The length of the beam maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 30.0 µT. What is the magnitude of the induced emf between the ends of the beam (in mV)?
Answer:14.2 mV
Explanation:
Given
Length of steel beam [tex]L=17\ m[/tex]
Magnetic field [tex]B=30\ \muT[/tex]
speed of beam [tex]v=28\ m/s[/tex]
Now consider as beam slides it encloses an area of rectangle of
width [tex]w=17\ m and length L'=v\times t [/tex]
Area [tex]A=17v\cdot t[/tex]
and Induced EMF is [tex]V=-\frac{d(BA)}{dt}[/tex]
[tex]V=-B\frac{d(17vt)}{dt}[/tex]
[tex]V=-17Bv[/tex]
[tex]V=-17\times 30\times 10^{-6}\times 28[/tex]
[tex]V=-14.2\ mV[/tex]
Magnitude of EMF[tex]=14.2\ mV[/tex]
A. When light of wavelength 250 nm is incident on a metal surface, the maximum speed of the photoelectrons is 4.0 × 105 m/s, what is the work function of the metal in electron volts? [10 marks] B. Assume that a 100-W light bulb gives off 2.50% of its energy as visible light of wavelength 500 nm. How many photons of visible light are given off in 1.0 min? Page 7 of 7 C. A small blood vessel near the skin surface has a radius of 10 μm, a length of 1 μm and the pressure drop along the blood vessel is 2.50 Pa. The viscosity of blood is 0.0027 Pa.s. i. What is the volume flow rate of blood through this blood vessel? What is the velocity of blood flow? [6 marks] ii. Vasodilation causes the radius of this blood vessel to increase to 12 μm, while leaving the pressure drop along the vessel unchanged? What is the volume flow rate through this blood vessel now? What is the velocity of blood flow? D. Many radioisotopes have important industrial, medical, and research applications. One of these is Cobalt-60, which has a half-life of 5.20 years and decays by the emission of a beta particle (energy 0.31 MeV) and two gamma photons (energies 1.17 MeV and 1.33 MeV). A scientist wishes to prepare a Cobalt-60 sealed source that will have an activity of at least 10 Ci after 30 months of use. i. What minimum initial mass of Cobalt-60 is required? [6 marks] ii. At what rate will the source emit energy after 30 months? [4 marks] E. The radioactive isotope Gold-198 has a half-life of 64.80 hrs. A sample containing this isotope has an initial activity of 40.0 μCi. Calculate the number of nuclei that will decay in the time interval from 10 hrs to 12 hrs. [10 marks]
Answer:
Explanation:
Part A
- Work function ( ∅ ) is the minimum energy required by the photon to knock an electron out of the metal surface. That is the portion of energy of a photon transferred to an electron so that it can escape the metal.
- We can mathematically express it:
∅ = Ep - Ek
= [tex]\frac{h*c}{lamba} - 0.5*m_e*v^2_e\\[/tex]
Where,
Planck's constant ( h ) = 6.6261*10^-34
Speed of light ( c ) = 3*10^8 m/s
mass of an electron ( m_e ) = 9.1094*10^-31 kg
Given:-
Incident light's wavelength ( λ ) = 250*10^-9 m
The maximum speed o electron ( v_e ) = 4*10^5 m/s
Solution:-
- Plug the values into the expression derived before:
∅ = [tex]\frac{(6.6261*10^-^3^4)*(3*10^8)}{250*10^-^9} \\\\[/tex]
∅ = [tex]7.22257*10^-^1^9 J * \frac{6.242*10^1^8 eV}{J}[/tex]
∅ = 4.508 eV ... Answer
Part B
- 2.5% of the energy emitted by a 100-W light bulb was visible light with wavelength ( λ ) = 500*10^-9 m. In 1.0 min the amount of energy harbored by a stream of photons are:
[tex]E_p = n_p*\frac{h*c}{lambda} = P*t*e[/tex]
Where,
Planck's constant ( h ) = 6.6261*10^-34
Speed of light ( c ) = 3*10^8 m/s
Given:-
visible light's wavelength ( λ ) = 500*10^-9 m
Power of light bulb ( P ) = 100 W
Time taken ( t ) = 1.0 min = 60 s
Portion of energy as light ( e ) = 0.025
Solution:-
- Plug the values into the expression derived before:
[tex]n_p = \frac{(lambda)*(P)*(t)*(e)}{h*c} \\\\n_p = \frac{(500*10^-^9)*(100)*(60)*(0.025)}{(6.6261*10^-^3^4)*(3*10^8)} \\[/tex]
n_p = 3.773 * 10^20 ... Answer
Part C
- A blood vessel of radius ( r ) with length ( L ) carries blood with viscosity ( μ ). The pressure drop ( ΔP ) in the blood vessel was witnessed.
- Pressure loss ( ΔP ) in a cylindrical blood vessel is given by the Darcy's equation given below:
[tex]\frac{dP}{p} = f*\frac{L}{D}*\frac{v^2}{2}[/tex]
Where,
ρ: Density of blood
f: Friction factor
D: Diameter of vessel
v: Average velocity
- The friction factor is a function of Reynolds number and relative roughness of blood vessel. We will assume the blood vessel to be smooth, round and the flow to be laminar ( later verified ).
- The flow rate ( Q ) in a smooth blood vessel subjected to laminar flow conditions is given by the Poiseuille's Law. The law states:
[tex]Q = \frac{\pi*dP*r^4 }{8*u*L}[/tex]
- The velocity ( v ) in a circular tube is given by the following relation:
[tex]v = \frac{Q}{\pi*r^2 }[/tex]
Given:-
dP ( Pressure loss ) = 2.5 Pa
radius of vessel ( r ) = 10μm = 10*10^-6 m
viscosity of blood ( μ ) = 0.0027 Pa.s
Length of vessel ( L ) = 1μm = 10^-6 m
Solution:-
- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood in the vessel:
[tex]Q = \frac{\pi*(2.5)*(10*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]
Q = 3.6361*10^-12 m^3 / s
- The corresponding velocity ( v ) of the blood flow would be:
[tex]v = \frac{3.6361*10^-^1^2}{\pi*(10*10^-^6)^2 }[/tex]
v = 0.01157 m/s
- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood blood in the enlarged vessel ( r = 12 μm = 10*10^-6 m ) :
[tex]Q = \frac{\pi*(2.5)*(12*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]
Q = 7.54*10^-12 m^3 / s
- The corresponding velocity ( v ) of the blood flow would be:
[tex]v = \frac{7.53982*10^-^1^2}{\pi*(12*10^-^6)^2 }[/tex]
v = 0.01666 m/s
Part D
- A radioactive isotope of Cobalt ( Co - 60 ) undegoes Beta decay ( 0.31 MeV ) and emits two gamma rays of energy ( 1.17 & 1.33 ) MeV.
- The radioactive decay for the ( Cobalt - 60 ) can be expressed in form of an equation:
[tex]_2_7Co ( 60 ) ---> _2_8 Ni ( 60 ) + e^- + v_e^- + gamma[/tex]
- The half life ( T_1/2 ) of the Co-60 can be used to determine the decay constant ( λ ):
λ = [tex]lambda = \frac{Ln(2)}{T_1/2}\\[/tex]
Where,
T_1/2 = 5.2 yrs = 1.68*10^8 s
Hence, the decay constant is
λ = [tex]\frac{Ln ( 2 ) }{1.68*10^8}[/tex] = 4.22*10^-9 s^-1
- The activity ( A ) of any radioactive isotope is function of time ( t ) defined by negative exponential distribution:
[tex]A = A_o*e^(^-^l^a^m^b^d^a^*^t^)[/tex]
Where,
A_o: The initial activity ( Bq )
- The activity of the radioactive isotope Co-60 was A = 10 Ci after t = 30 months. The initial activity ( A_o ) can be determined:
[tex]A_o= \frac{A}{e^(^-^l^a^m^b^d^a^*^t^)} \\\\A_o= \frac{(10Ci)*(3.7*10^1^0 Bq/Ci)}{e^(^-^4^.^2^2^*^1^0^-^9*^7^.^8^8^*^1^0^7^)} \\[/tex]
A_o = 5.016 * 10^11 Bq
- The initial number of nuclei in the sample ( N_o ) is given by:
[tex]N_o = (A_o) / (lambda)\\\\N_o = \frac{5.016*10^1^1}{4.22*10^-^9} = 1.22*10^2^2[/tex]
- The initial mass of Co-60 used as a sample can be determined:
[tex]m_o = M_r*N_o\\\\m_o = (59.933822u)*(1.66*10^-^2^7)*(1.22*10^2^2)\\[/tex]
m_o = 12.2 * 10^-6 kg ... Answer
- The total energy ( E ) released from the beta decay transformation:
E = E(β) + E(γ1) + E(γ2) = 0.31 + 1.17 + 1.33 = 2.81 MeV
- The rate at which the source emits energy after 30 months:
P = E*A = [tex]( 2.81 MeV * \frac{1.6*10^-^1^3 J}{MeV} ) * ( 10 Ci * \frac{3.7*10^1^0 Bq}{Ci} )[/tex]
P = 0.166 W .. Answer
In an oscillating LC circuit, L = 4.24 mH and C = 3.02 μF. At t = 0 the charge on the capacitor is zero and the current is 2.38 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?
Answer:
a) 2.693*10^-4 C
b) 8.875*10^-5 s
c) 2.96 W
Explanation:
Given that
Inductance of the circuit, L = 4.24 mH
Capacitance of the circuit, C = 3.02 μF
Current in the circuit, I = 2.38 A
See attachment for calculations
Answer:
a) 0.269 mC
b) 0.355 ms
c) 1.39W
Explanation:
a) To find the charge off the capacitor you start by using the following expression for the charge in the capacitor:
[tex]q=Qsin(\omega t)[/tex]
next, you calculate the current I by using the derivative of q:
[tex]I=\frac{dq}{dt}=Q\omega cos(\omega t)\\\\for \ t= 0:\\\\I=Q\omega\\\\Q=\frac{I}{\omega}\\\\\omega=\frac{1}{\sqrt{LC}}\\\\Q=I\sqrt{LC}[/tex] ( 1 )
L: inductance = 4.24*10^{-3}H
C: capacitance = 3.02*10^{-6}F
I: current = 2.38 A
you replace the values of the parameters in (1):
[tex]Q=(2.38A)(\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)})=2.69*10^{-4}C=0.269mC[/tex]
b) to find the time t you use the following formula for the energy of the capacitor:
[tex]u_c=\frac{q^2}{2C}=\frac{Q^2sin^2(\omega t)}{2C}[/tex]
the maximum storage energy in the capacitor is obtained by derivating the energy:
[tex]\frac{du_c}{dt}=\frac{2\omega Q^2sin(\omega t)cos(\omega t)}{2C}=0\\\\\frac{du_c}{dt}=\frac{\omega Q^2 sin(2\omega t)}{2C}=0\\\\sin(2\omega t)=0\\\\2\omega t= 2\pi\\\\t=\frac{\pi}{\omega}=\pi\sqrt{LC}=\pi\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)}=3.55*10^{-4}s=0.355\ ms[/tex]
hence, the time is 0.355 ms
c) The greatest rate is obtained for duc/dt evaluated in t=0.355 ms:
[tex]\frac{du_c}{dt}=\frac{2Q^2sin(2\frac{t}{\sqrt{LC}})}{\sqrt{LC}}[/tex]
[tex]\frac{du_c}{dt}=\frac{(2.69*10^{-4}C)^2sin(2\frac{3.55*10^{-4}s}{\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}})}{2(3.02*10^{-6}C)\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}}=-1.39W[/tex]
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
USE THE ABOVE INFORMATION TO ANSWER QUESTION 3 TO 6
The focal length of a convex mirror is 10 cm and the object distance is 20
Determine the image height if the peg place at the object distance of 5 cm from the concave mirror
4. Determine the image height if the peg place at the object distance of 15 cm from the concave mirror
5. Determine the image height if the peg place at the object distance of 20 cm from the concave mirror
6. Determine the image height if the peg place at the object distance of 30 cm from the concave mirror
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
how much force is required to accelerate an 1800 kg truck 3 m/s
Answer:
5,400 Newtons.
Explanation:
Force = 1,800 kg * 3 m/s^2 = 5,400 kg*m/s^2 = 5,400 N
A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor?
Answer:
A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor?
The median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.
Explanation:
Based on the scenario described in the question, the median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.
When Brett and Will ride the carousel, Brett always selects a horse on the outside row, whereas Will prefers the row closest to the center. These rows are 19 ft 1 in.19 ft 1 in. and 11 ft 11 in.11 ft 11 in. from the center, respectively. The angular speed of the carousel is 2.72.7 revolutions per minute. What is the difference, in miles per hour, in the linear speeds of Brett and Will?
Answer:
the difference, in miles per hour, in the linear speeds of Brett and Will;
∆v = 1.38 mph
Explanation:
Given;
Angular speed w = 2.7 revolutions per minute
Converting to revolutions per hour
w = 2.7 × 60 revolutions per hour
w = 162 rev/hour
Linear speed v = angular speed × 2πr
the difference, in miles per hour, in the linear speeds of Brett and Will;
∆v = w × 2π(r1 - r2)
r1 = Brett radius in miles
r2 = Will radius in miles
r1 = 19ft 1in = (19×12 + 1) = 229 in
r1 = 229 × 1.57828283 × 10^-5 miles
r2 = 11 ft 11 in = (11×12 + 11) = 143 in
r2 = 143 × 1.57828283 × 10^-5 miles
Substituting the values;
∆v = 162 × 2π × (229-143)×1.57828283 × 10^-5 mph
∆v = 1.38 mph
g A fiber optic is made by covering a thin fiber core material with a cladding material. An optical fiber works on the principle of total internal reflection, from this we can conclude that: A) The refractive index of the core material is less than the refractive index of the cladding material. B) The refractive index of the cladding material is less than the refractive index of the core material. C) They must have the same refractive index. D) It does not matter what the refractive index is. E) The refractive index of the cladding must be exactly twice that of the core.
Answer:(b)
Explanation:
Optical fiber works on the principle of total internal reflection . Total internal reflection occurs when the following two conditions are met
(1) When the angle of incidence of light is greater than the critical angle and,
(2) When a ray of light travels from a optically denser medium to optically rarer medium .
We know that light bend away from normal when it travels from denser to rarer medium. If we we choose such an angle that the refraction angle is [tex]90^{\circ}[/tex] then the incident angle is called critical angle.
So, the refractive index of cladding must be must be less than the refractive index of core material.
A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λA, λB, and λC). She observes three distinct first-order spectral lines at the following respective angles θm (where m denotes order). θ1 = 12.9°, θ1 = 14.2°, θ1 = 15.0° (a) If the grating has 3,680 grooves per centimeter, what wavelength (in nm) describes each of these spectral lines?
Answer:
Explanation:
Given
[tex]N=3680 cm^{-1}[/tex]
therefore slit spacing [tex]d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm[/tex]
since [tex]d\sin \theta =n\lambda [/tex]
for [tex]n=1[/tex]
[tex]d\sin \theta =\lambda [/tex]
Now,at [tex]\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm[/tex]
at [tex]\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm[/tex]
at [tex]\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm[/tex]
White light, with a uniform intensity across the visible wavelength 400 – 690 nm, is perpendicularly incident on a water film, having index of refraction n2 = 1.33 and a thickness of 320 nm, that is suspended in air. At what wavelength is the light reflected by the film brightest to an observer? Leave your final answer in nanometers.
Answer:
Wavelength at which the light reflected by the film is brightest = 567.5 nm
Explanation:
We are given;
index of refraction n2 = 1.33
Thickness;(t) = 320 nm
Now the wavelength at which the light reflected by the film is brightest is gotten from the formula for path difference in critical interference as;
Path difference = (m + ½)(λ/n)
Where;
path difference = 2 x thickness = 2(320) = 640 nm
λ = Wavelength at which the light reflected by the film is brightest
n is Refractive index
m is an integer = 0,1,2,3...
Thus; at m = 0;
We have;
640 = (0 + ½)(λ/1.33)
640 = (λ/2.66)
λ = 640 x 2.66
λ = 1702.4 nm
at m = 1;
We have;
640 = (1 + ½)(λ/1.33)
640 = (3/2)(λ/1.33)
λ = 640 x 1.33 x 2/3
λ = 567.5 nm
at m = 2;
We have;
640 = (2 + ½)(λ/1.33)
640 = (5/2)(λ/1.33)
λ = 640 x 2 x 1.33/5
λ = 340.5 nm
Since we are told that the wavelength is between 400 – 690 nm.
Thus, the wavelength at which the light reflected by the film is brightest is the higher value gotten that is between 400nm and 690nm.
Thus, Wavelength at which the light reflected by the film is brightest = 567.5 nm
The light reflected by the water film will appear brightest to an observer at a wavelength of approximately 854.4 nm.
Explanation:The wavelength of light in a medium is given by λn = λ/ʼn, where λ is the wavelength in vacuum and ʼn is the medium's index of refraction. In water, which has an index of refraction of n = 1.33, the range of visible wavelengths is 285 to 570 nm. When light is incident on the water film, it will reflect predominantly at the wavelengths where the film's thickness is an integer multiple of half the wavelength. The brightest reflection will occur when the film's thickness is such that the reflected wavelength is at its maximum in the range of visible light.
To find the wavelength of the brightest reflection, we can use the equation λn = 2nt, where λn is the wavelength in vacuum, n is the refractive index of the film, and t is the thickness of the film. Given the refractive index n2 = 1.33 and the thickness t = 320 nm, we can solve for λn.
Plugging in the values:
λn = 2nt = 2(1.33)(320 nm) ≈ 854.4 nm
Therefore, the wavelength of the light reflected by the film that appears brightest to an observer is approximately 854.4 nm.
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The electric field at the center of a ring of charge is zero. At very large distances from the center of the ring along the ring's axis the electric field goes to zero. Find the distance from the center of the ring along the axis (perpendicular to the plane containing the ring) at which the magnitude of the electric field is a maximum. The radius of the ring is 6.58 cm and the total charge on the ring is 8.87E-6 C.
Answer:
z = 1.16m
Explanation:
The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:
[tex]E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}[/tex]
z: distance to the plane of the ring
r: radius of the ring
Q: charge of the ring
you have that:
E_{z->0} = 0
E_{z->∞} = 0
To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:
[tex]\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}\\\\[/tex]
you take the positive value:
[tex]z^2=\frac{2+3.464}{4}=1.366\\\\z=1.16m[/tex]
hence, the distance in which the magnitude if the electric field is maximum is 1.16m
A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg⋅m². You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is 73.0 kg. Since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip.
What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?
Answer:
0.8024
Explanation:
From the given question; we can say that the angular momentum of the system is conserved if the net torque is is zero.
So; [tex]I_o \omega _o = I_2 \omega_2[/tex]
At the closest distance ; the friction is :
[tex]f_s = \mu_s (mg)[/tex]
According to Newton's Law:
F = ma
F = mrω²
From conservation of momentum:
[tex]I_o \omega _o = I_2 \omega_2[/tex]
[tex]\omega_2= \frac { I_o \omega _o}{ I_2 }[/tex]
[tex]\omega_2=( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })* \omega_o[/tex]
However ; since the static friction is producing the centripetal force :
[tex]\mu_s (mg) = mr \omega_2^2[/tex]
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
The coefficient of static friction between the bottoms of your feet and the surface of the turntable can now be calculated by using the formula :
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
= [tex]( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })^2 * \frac{ \omega^2_o *r }{g}[/tex]
= [tex][\frac{1200+(73(6)^2)}{((1200)+(73)(3)^2)} ]^2*[\frac{(\frac{\pi}{4})^2(3)} {9.8}][/tex]
= 0.8024
The friction force between the bottom of the feet and the surface of the
turntable balance the centrifugal force due to rotation.
The coefficient static friction is approximately 0.802Reasons:
Radius of the turntable, r₀ = 6.00 m
The period of rotation of the turntable, T₀ = 8.00s
Moment of inertia of the turntable, [tex]I_t[/tex] = 1,200 kg·m²
Mass of the person, m = 73.0 kg
The point at which the feet starts to slip, r₁ = 3.00 m
Required:
The friction between the bottom of the feet and the surface of the turntable.
Solution:
The moment of inertia of the person and the turntable combined, J, can be
found by considering the person as a point mass.
Therefore;
At the rim, J₀ = [tex]I_t[/tex] + m·r² = 1,200 + 73×6² = 3828
J₀ = 3,828 kg·m²
At 3.00 m from the center, J₁ = [tex]I_t[/tex] + m·r² = 1,200 + 73×3² = 1,857
J₁ = 1,857 kg·m²
Angular momentum, L = J·ω
Whereby the angular momentum is conserved, we have;
L₀ = L₁
J₀·ω₀ = J₁·ω₁
[tex]\displaystyle \omega_1 = \mathbf{ \frac{J_0 \cdot \omega_0}{J_1}}[/tex]
Which gives;
To remain at equilibrium, the friction force, [tex]F_f[/tex] = The centrifugal force, [tex]F_c[/tex]
[tex]F_f =\mu_s \cdot W = \mu_s \cdot m \cdot g[/tex][tex]\displaystyle F_c = \frac{m \cdot v^2}{r} = \frac{m \cdot (\omega \cdot r)^2}{r} = m \cdot r \cdot \omega^2[/tex]
Where;
[tex]\mu_s[/tex] = The coefficient of static friction
g = Acceleration due to gravity which is approximately 9.81 m/s²
r = The radius of rotation
ω = The angular speed
[tex]F_f[/tex] = [tex]F_c[/tex]
Therefore;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{m \cdot r \cdot \omega^2}[/tex]Therefore, at 3.00 m from the center of the turntable, we have;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{ m \cdot r_1 \cdot \omega_1^2}[/tex]
[tex]\displaystyle\mu_s \cdot g = r_1 \cdot \omega_1^2[/tex]
[tex]\displaystyle\mu_s = \frac{r_1 \cdot \omega_1^2}{g} = \mathbf{\frac{r_1 \cdot \left(\displaystyle \frac{J_0 \cdot \omega_0}{J_1}\right)^2}{g}}[/tex]
Which gives;
[tex]\displaystyle\mu_s =\frac{3 \times \left(\displaystyle \frac{3,828 \times \frac{2 \cdot \pi}{8} }{1,857}\right)^2}{9.81} \approx \mathbf{0.802}[/tex]
The coefficient of static friction between the bottom of the feet and the surface of the turntable, at 3.00 m from the center, [tex]\mu_s[/tex] ≈ 0.802.
Learn more about the conservation of angular momentum here:
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Please help me with this question :
A soap bubble has the index of refraction n = 1.33. There is air both inside and outside
the bubble.
a/ What wavelength (in air) of visible light is most strungly reflected from a point on a soap bubble where
its wall is 290 nm thick? To what color does this correspond?
b/ Repeat part a/ for a wall thickness of 340 nm. What is your observation about the two answers a/ and b/?
Answer:
514.27 ( wavelength )the color is green
602.93 nm ( orange color )the observation is that there is a change of visible color
Explanation:
A) wavelength of visible light that is most strongly reflected from a point on a soap
refraction n = 1.33
wall thickness (t) = 290 nm
2nt = (2m +1 ) ∝/2 -----equation 1
note when m = 0
therefore ∝ = 4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this
when m = 1
equation 1 becomes
∝ = 4nt/3 =( 4 * 1.33 * 290) / 3 = 1542.8 / 3 = 514.27 ( wavelength )
the color is green
B) the wavelength when the wall thickness is 340 nm
∝ = 4nt / 2m +1
where m = 1
∝ = (4 * 1.33 * 340 ) / 3 = 1808.8 / 3 = 602.93 nm ( orange color )
the observation is that there is a change of visible color
Final answer:
The wavelength of visible light most strongly reflected for a soap bubble wall thickness of 290 nm and index of refraction of 1.33 is 770 nm, corresponding to the red color. For a thickness of 340 nm, the wavelength is 904 nm, which falls outside the visible spectrum, in the infrared range. As the thickness increases, the reflected wavelength shifts toward longer wavelengths.
Explanation:
The phenomenon described in the question is known as thin film interference, which is a physical effect that occurs when light waves reflected off the top and bottom surfaces of a thin film interfere with each other. To find the wavelength (λ) of visible light that is most strongly reflected, we use the formula for constructive interference in thin films:
2nt = mλ,
where n is the index of refraction of the film, t is the thickness of the film, and m is the order of the fringe (which is an integer).
For part (a), we have a soap bubble thickness of 290 nm and an index of refraction n = 1.33. Assuming the light is perpendicular to the surface (m = 1 for the first order of constructive interference), we calculate the wavelength using the given formula as follows:
λ = 2nt/m = 2 * 1.33 * 290 nm / 1 = 770 nm.
The wavelength of 770 nm corresponds to red light in the visible spectrum.
For part (b), with a wall thickness of 340 nm, we calculate the wavelength in a similar fashion:
λ = 2 * 1.33 * 340 nm / 1 = 904 nm.
However, a wavelength of 904 nm falls outside the visible spectrum and cannot be seen as a color. It is in the infrared range.
Comparing answers from (a) and (b), we observe that as the thickness of the soap bubble increases, the wavelength of light most strongly reflected shifts towards the longer wavelengths, moving out of the visible range into the infrared.
1. Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizontally, and close enough so that both of them fall in the beam cross section. By shifting vertically the screen, each slits can be illuminated independently, allowing the other to be disregarded. On an observation screen positioned 1m further from the opaque screen, the diffraction patterns from the independent slit illumination were found identical, with minima 6.5 mm apart. The interference pattern maxima from simultaneous illumination of both slits were 0.53 mm apart. What are the double slit characteristics (slit width, slit separation)
To find the double-slit characteristics of width and separation, one must use the data from a single-slit diffraction pattern to calculate the slit width and then the interference pattern from both slits to find the slit separation.
The student's question is about the characteristics of a double-slit diffraction pattern produced when a laser beam with a specific wavelength is projected through two slits. To deduce the slit width and separation, we use the information provided about the diffraction and interference patterns observed.
Firstly, we use the minima separation from the single slit diffraction to find the slit width using the diffraction formula d sin(θ) = mλ, where d is the slit width, m is an integer denoting the order of the minima, λ is the wavelength, and θ is the angle of the minima. Since the screen is 1 meter away and the minima are 6.5 mm apart, we can approximate sin(θ) as the ratio of minima separation to the distance to the screen to solve for the slit width d.
Next, we apply the formula for double-slit interference, d sin(θ) = nλ (where d is now the slit separation, n is an integer denoting the order of the maxima) to the observed interference maxima separation to calculate the slit separation.
Slit separation: [tex]\(97.23 \times 10^{-6}\)[/tex] m. Slit width: [tex]\(1194.34 \times 10^{-6}\)[/tex] m. Calculated using interference formula and provided data.
To solve this problem, we can use the double-slit interference formula:
[tex]\[ \Delta y = \frac{\lambda L}{d} \][/tex]
where:
- [tex]\( \Delta y \)[/tex] is the distance between adjacent maxima or minima on the observation screen,
- [tex]\( \lambda \)[/tex] is the wavelength of the laser beam,
- [tex]\( L \)[/tex] is the distance between the screen and the slits, and
- [tex]\( d \)[/tex] is the slit separation.
Given:
- Wavelength [tex]\( \lambda = 632.8 \)[/tex] nm = [tex]\( 632.8 \times 10^{-9} \)[/tex] m,
- Distance between the screen and the slits [tex]\( L = 1 \)[/tex] m = 1 m,
- Distance between adjacent minima [tex]\( \Delta y_{\text{min}} = 6.5 \)[/tex] mm = [tex]\( 6.5 \times 10^{-3} \)[/tex] m, and
- Distance between adjacent maxima [tex]\( \Delta y_{\text{max}} = 0.53 \)[/tex] mm = [tex]\( 0.53 \times 10^{-3} \)[/tex] m.
First, let's find the slit separation [tex]\( d \)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} \][/tex]
Then, let's find the slit width [tex]\( w \)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} \][/tex]
Let's calculate these values.
First, let's calculate the slit separation [tex]\(d\)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{6.5 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ d \approx \frac{(632.8)(1)}{6.5} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ d \approx 97.23 \times 10^{-6} \, \text{m} \][/tex]
Now, let's calculate the slit width [tex]\(w\)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{0.53 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ w \approx \frac{(632.8)(1)}{0.53} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ w \approx 1194.34 \times 10^{-6} \, \text{m} \][/tex]
So, the slit separation [tex]\(d\)[/tex] is approximately [tex]\(97.23 \times 10^{-6}\)[/tex] m, and the slit width [tex]\(w\)[/tex] is approximately [tex]\(1194.34 \times 10^{-6}\)[/tex] m.
How has God designed sound waves so that a person is able to hear all the different instruments in a musical piece at the same time?
Answer: when the wave encounters something, it can bounce (reflection) or be bent (refraction). In fact, you can "trap" waves by making them bounce back and forth between two or more surfaces. Musical instruments take advantage of this; they produce pitches by trapping sound waves.
Explanation: Any bunch of sound waves will produce some sort of noise. But to be a tone - a sound with a particular pitch - a group of sound waves has to be very regular, all exactly the same distance apart. That's why we can talk about the frequency and wavelength of tones.
Final answer:
The human ear hears different instruments in music at once due to the superposition of sound waves and the ear's ability to process overlapping sounds, determining their pitch, loudness, and direction.
Explanation:
The human ear can hear all the different instruments in a musical piece simultaneously because of the principle of superposition of sound waves. Sound waves combine without affecting each other, so multiple sounds can reach our ears at the same time. Additionally, our ears and brain are capable of processing and interpreting these overlapping waves, enabling us to differentiate between the various sources of sound.
This is partly due to the design of the ear, which is sensitive to a wide range of frequencies, and the ear's ability to discern pitch, loudness, and direction of sound. The cochlea inside the ear uses temporal and place theories to help us perceive pitch, which allows us to enjoy the complexities of music and speech.
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the distance of point charge on the line joining the charge B. Where the resultant electric field is zero
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
Water is completely filling black metallic vessel having cubic form and thin walls. The mass of water is 1 kg and initial temperature 50 C. This cubic vessel is placed into black cavity whose walls are kept at the constant temperature of 0 C. Find time t it takes for the water to cool down to 10 C. Water specific heat is 4200 (J/kg K-1 .) After you find an answer, estimate whether this radiation mechanism would be responsible for the real-life water cooling and what other mechanisms might be in play.
Answer:
Check the attached image
Explanation:
To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.
Kindly check the attached image below for the step by step explanation to the question.