Answer:
The uncertainty in momentum is 5.25x 10^25Jsm
Explanation:
We know that
h bar = h/2π
So
1.05x 10^34=h/2pπ
h=1.05x 10^ 34(2π)=6.597x 10^-34Js
dp=(6.597x10^-34/4pπ)/(1x10^-10)
=5.25x10^-25 Jsm
The element sodium can emit light at two wavelengths, λ1 = 588.9950 nm and λ2 = 589.5924 nm. Light in sodium is being used in a Michelson interferometer. Through what distance must mirror M 2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?
Answer:
The distance is [tex]d = 0.00029065 \ m[/tex]
Explanation:
From the question we are told that
The first wavelength is [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]
The second wavelength is [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]
The difference in the fringe pattern is n = 1.0
Generally the equation defining the effect of the movement of the mirror M 2 in a Michelson interferometer is mathematically represented as
[tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]
Here d is the mirror M 2 must be moved
substituting values
[tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]
[tex]d = 0.00029065 \ m[/tex]
The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.
Answer:
1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz
Explanation:
1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore
L = λ / 2
λ = 2L
λ = 2 0.8
λ = 1.6 m
wavelength and frequency are related to the speed of sound (v = 343 m / s)
v =λ f
f = v / λ
f = 343 / 1.6
f = 214 Hz
2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced
λ' = 2 L ’
as L ’<L₀
λ' <λ₀
f = v / λ'
f' > fo
the correct answer is c
3) in this case the length is L = 0.40 m
λ = 2 0.4 = 0.8 m
f = 343 / 0.8
f = 428 Hz
4) the different harmonics are described by the expression
λ = 2L / n n = 1, 2, 3
λ₂ = L
f₂ = 343 / 0.8
f₂ = 428 Hz
λ₃ = 2 0.8 / 3
λ₃ = 0.533 m
f₃ = 343 / 0.533
f₃ = 643 Hz
4,1) as we have two maximums at the ends, all integer multiples are present
the answer is C
E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested
in this pipe there is a maximum in the open part and a node in the closed part, so the expression
L = λ / 4
L = 1.6 / 4
L = 0.4 m
the answer is C
F) in this type of pipe the general expression is
λ = 4L / n n = 1, 3, 5 (2n + 1)
therefore only odd values can produce standing waves
λ₃ = 4L / 3
λ₃ = 4 0.4 / 3
λ₃ = 0.533
f₃ = 343 / 0.533
f₃ = 643 Hz
A wire of 0.50m length is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.98T. The vurrent in the wire is 2.0A in the direction shown. What is the mass of the wire if the current and the magnetic field are sufficient to remove the tension in the supporting leads?
Answer:
0.1 kg or 100 g
Explanation:
The length of the wire = 0.5 m
the field magnitude = 0.98 T
the current through the wire = 2.0 A
magnetic force due to a wire carrying current is
F = [tex]IlB[/tex]
where
F is the force
[tex]I[/tex] is the current = 2 A
[tex]l[/tex] is the length of the wire
B is the magnetic field strength
Substituting, we have
F = 2 x 0.5 x 0.98 = 0.98 N
This force balances the weight of the mass
weight = mg
where m is the mass of the wire
g is acceleration due to gravity = 9.81 m/s^2
therefore, weight = m x 9.81 = 9.81m
equating this weight with the force, we have
0.98 = 9.81m
m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g
Answer:
100 g
Explanation:
How long will it take a spacecraft travelling at 99% the speed of light (gamma = 7) to reach
the star Sirius which is 8.6 light-years away according to people on Earth ? How long will it
take according to the crew of the ship?
Answer:
The time taken is [tex]t = 2.739 *10^{8} \ s[/tex]
Explanation:
From the question we are told that
The speed of the spacecraft is [tex]v = 0.99c[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]
The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]
[tex]t = 2.739 *10^{8} \ s[/tex]
A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on the
Given that,
Current = 4 A
Sides of triangle = 50.0 cm, 120 cm and 130 cm
Magnetic field = 75.0 mT
Distance = 130 cm
We need to calculate the angle α
Using cosine law
[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]
[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]
[tex]\alpha=\cos^{-1}(0.3846)[/tex]
[tex]\alpha=67.38^{\circ}[/tex]
We need to calculate the angle β
Using cosine law
[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]
[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]
[tex]\beta=\cos^{-1}(0.923)[/tex]
[tex]\beta=22.63^{\circ}[/tex]
We need to calculate the force on 130 cm side
Using formula of force
[tex]F_{130}=ILB\sin\theta[/tex]
[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]
[tex]F_{130}=0[/tex]
We need to calculate the force on 120 cm side
Using formula of force
[tex]F_{120}=ILB\sin\beta[/tex]
[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]
[tex]F_{120}=0.1385\ N[/tex]
The direction of force is out of page.
We need to calculate the force on 50 cm side
Using formula of force
[tex]F_{50}=ILB\sin\alpha[/tex]
[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]
[tex]F_{50}=0.1385\ N[/tex]
The direction of force is into page.
Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.
a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.
b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.
c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.
Given the following data:
Current = 4.00 Amperes.Magnetic field strength = 75.0 mT = [tex]7.5 \times 20^{-3}\;T[/tex]Length = 130 cm to m = 1.3 mHypotenuse = 130 cmOpposite side = 120 cmAdjacent side = 50 cmLet us assume the current is flowing in a counterclockwise direction in the right-angle triangle.
First of all, we would determine the angles by using cosine rule:
[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]
[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]
a. To the determine the magnitude of the magnetic force on the 130 cm side:
Mathematically, the force acting on a current in a magnetic field is given by the formula:
[tex]F = BILsin\theta[/tex]
Where:
B is the magnetic field strength.I is the current flowing through a conductor.L is the length of conductor.[tex]\theta[/tex] is the angle between a conductor and the magnetic field.Substituting the given parameters into the formula, we have;
[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]
b. To the determine the magnitude of the magnetic force on the 120 cm side:
[tex]F_{120}=BILsin\beta[/tex]
[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]
c. To the determine the magnitude of the magnetic force on the 50 cm side:
[tex]F_{50}=BILsin\alpha[/tex]
[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]
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find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popped up vertically velocity of 32 m/s
Answer:
whats the formula
Explanation:
how can you relazie a perfect balck body in pratice
The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.
Answer:
valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.
I hope this helps
An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA. Part A What is the resistance RR of the inductor
i
CHECK COMPLETE QUESTION BELOW
inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.
Part A)What is the resistance RR of the inductor
PART B) what is inductance L of the conductor
Answer:
A)R=1818.18 ohms
B)L=1.0446H
Explanation:
We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.
There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,
.
a)A)What is the resistance RR of the inductor?
The current flowing into RL circuit can be calculated using below expresion
i=ε/R[1-e⁻(R/L)t]
at t=∞ there is maximum current
i(max)= ε/R
Where ε emf of the battery
R is the resistance
R=ε/i(max)
= 12V/(6.60*10⁻³A)
R=1818.18 ohms
Therefore, the resistance R=1818.18 ohms
b)what is inductance L of the conductor?
i(t=0.80ms and 4.96mA
RT/L = ⁻ln[1- 1/t(max)]
Making L subject of formula we have
L=-RT/ln[1-i/i(max)]
If we substitute the values into the above expresion we have
L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]
L=1.0446H
Therefore, the inductor L=1.0446H
I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)
Solve for a
Answer:
a. solve for a
[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]
I hope I helped you ^_^
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Answer:
b. The current in the loop always flows in a counterclockwise direction.
Explanation:
When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.
The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.
This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
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In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
A collector that has better efficiency in cold weather is the:
flat-plate collector due to reduced heat loss
evacuated tube collector due to its larger size
flat-plate collector due to the dark-colored coating
O evacuated tube collector due to reduced heat loss
Question 23 (1 point) Saved
One of the following is not found in Thermosyphon systems
o
Answer:
D. evacuated tube collector due to reduced heat loss
Explanation:
Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.
During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.
Trong thực nghiệm nhiễu xạ qua 1 khe hẹp, bề rộng khe là 2.10-5 m, màn quan sát đặt cách khe 0,5 m. Chiếu ánh sáng có bước sóng 480 nm xuyên qua khe và thấy có nền nhiễu xạ trên màn quan sát. Tính bề rộng của vân sáng liền kề vân sáng trung tâm.
Answer:
tyygggghgtyhyrdfgyyyhjjillbxsrfvgygvnjj
Explanation:
cffczhxucuxoyitxohvojcdivbjv ohxc
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?
Answer:
Time taken for 1 swing = 3.81 second
Explanation:
Given:
Time taken for 1 swing = 2.20 Sec
Find:
Time taken for 1 swing , when triple the length(T2)
Computation:
Time taken for 1 swing = 2π[√l/g]
2.20 = 2π[√l/g].......Eq1
Time taken for 1 swing , when triple the length (3L)
Time taken for 1 swing = 2π[√3l/g].......Eq2
Squaring and dividing the eq(1) by (2)
4.84 / T2² = 1 / 3
T2 = 3.81 second
Time taken for 1 swing = 3.81 second
Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Answer:
450 x10^-6 T
Explanation:
We know that the magnetic of each wire is derived from
ByB= uoi/2pir
Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)
= 0.2 x 10^ -3T
B2=
4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)
= 0.4 x 10^ -3T
So
(Bnet)² = (Bx)² + ( By)²
= (0.2² + 0.4²)mT
= 450 x10^-6T
The magnitude of magnetic field at the origin is required.
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
x = Location at x axis = 5 cm
y = Location at y axis = 4 cm
[tex]I_x[/tex] = Current at the x axis point = 50 A
[tex]I_y[/tex] = Current at the y axis point = 80 A
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]
Magnitude of the magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
Finding the net magnetic field using the Pythagoras theorem
[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
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If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
If Vector A is (6, 4) and Vector B is (-2, -1), what is A – B?
A. (8,5)
B. (4,5)
C. (4,3)
D. (8,3)
Answer:
I think the answer is A...I'm not sure
Explanation:
A=(6,4)
B=(-2,-1)
A-B=(6-(-2)),(4-(-1))
=(6+2),(4+1)
=(8,5)
Answer:
[tex]6-(-2)=[/tex]
[tex]6+2[/tex]
[tex]=8[/tex]
[tex]4-\left(-1\right)[/tex]
[tex]=4+1[/tex]
[tex]=5[/tex]
[tex](8,5)[/tex]
[tex]\textbf{OAmalOHopeO}[/tex]
How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?
Answer:
[tex]y = 0.0394 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.20 \ m[/tex]
The distance of separation of the slit is [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]
The wavelength of light is [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]
Generally the condition for constructive interference is
[tex]dsin\theta = n * \lambda[/tex]
=> [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]
here n = 1 because we are considering the central diffraction peak
=> [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]
=> [tex]\theta = 1.0274 ^o[/tex]
Generally the width of central diffraction peak on a screen is mathematically evaluated as
[tex]y = D tan (\theta )[/tex]
substituting values
[tex]y = 2.20 * tan (1.0274)[/tex]
[tex]y = 0.0394 \ m[/tex]
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 dB and 20 dB respectively
Answer:
The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
Explanation:
The intensity of sound is given by;
[tex]I(dB) = 10Log(\frac{I}{I_o} )[/tex]
where;
I is the intensity of the sound
I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²
The intensity of sound at a rock concert
[tex]120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2[/tex]
The intensity of sound of a whisper
[tex]20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2[/tex]
Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
When an LRC series circuit is driven at resonance, which of the following statements about the circuit are correct? (There may be more than one correct choice.)a) The impedance of the circuit has its minimum value.b) The inductive reactance and the capacitive reactance are exactly equal to each other.c) The impedance of the circuit is zero.d) The inductive reactance and the capacitive reactance are both zero.e) The impedance of the circuit has its maximum value.
Answer:
Answers a) and b) should be marked as correct.
Explanation:
Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:
[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]
as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.
Therefore, answers:
a) "The impedance of the circuit has its minimum value."
and
b) "The inductive reactance and the capacitive reactance are exactly equal to each other."
are correct answers.
(a) The impedance of the circuit has its minimum value.
(b) The inductive reactance and the capacitive reactance are exactly equal to each other
LRC series circuit consists of inductor, resistor and capacitor is series.
The impedance of the circuit is calculated as follows;
[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]
where;
R is the resistance[tex]X_C[/tex] is the capacitive reactance[tex]X_L[/tex] is the inductive reactanceThe impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.
[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]
Learn more here:https://brainly.com/question/17330472
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
16. If one body is positively charged and another body is negatively charged, free electrons tend to
O A. move from the negatively charged body to the positively charged body
O B. remain in the positively charged body
OC. move from the positively charged body to the negatively charged body
O D. remain in the negatively charged body
Answer:
Hey there!
The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body
Let me know if this helps :)
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
why is nut-cracker 2nd class lever?
2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......
A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an angle of 30 degrees with the horizontal. The worker believes that he can get the crate to the very top of the ramp by launching it at 5 m/s at the bottom and letting go. But friction is not neglible; the crate slides 1.6 m upthe ramp, stops, and slides back down.
Required:
a. Assuming that the friction force actingon the crate is constant, find its magnitude.
b. How fast is teh crate moving when it reachesthe bottom of the ramp?
Answer:
a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.
Explanation:
a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.
By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
Where:
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.
The equation is now simplified and the coefficient of friction is consequently cleared:
[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]
[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]
The final height of the crate is:
[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]
[tex]y_{2} = 0.8\,m[/tex]
If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:
[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\mu_{k} \approx 0.548[/tex]
Then, the magnitude of the friction force is:
[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:
[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]f = 55.851\,N[/tex]
The magnitude of the force of friction is 55.851 newtons.
b) The energy equation of the situation is:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]
Now, the final speed is cleared:
[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]
[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]
Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:
[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]
The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.