Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + [tex]\frac{1}{2}[/tex] ) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + [tex]\frac{1}{2}[/tex])v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + [tex]\frac{1}{2}[/tex])v / Δr
f = [tex]\frac{1}{2}[/tex](v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = [tex]\frac{1}{2}[/tex](343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz
a. Why don't we hear the sound of oscillation of second
pendulum?
Answer:
because the frequency is too low for humans to hear.
Explanation:
Answer:
we cannot hear the sound produced due to vibrations of a seconds' pendulum.
Explanation:
This is because the frequency of sound produced as a result of vibrations of seconds' pendulum is 0.5Hz0.5Hz which is infrasonic sound.
Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?
Answer:
[tex]0.0035\ \text{m}[/tex]
Explanation:
y = Distance from the center point
d = Separation between slits = 0.3 mm
D = Distance between slit and screen = 1.5 m
[tex]\lambda[/tex] = Wavelength = 700 nm
m = Order = 1
We have the relation
[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]
The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.54 s . His older sister pulls the spring a bit more than intended. She pulls the animal 32 cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. Part A If the animal does not hit anything on the way up, how far above its equilibrium position will it go
Answer:
the wooden animal will go 0.7068 m above its equilibrium
Explanation:
Given the data in the question;
mass of wooden animal m = 120 g = 0.12 kg
the animal oscillates up and down, T = 0.54 s
older sister pulls the animal 32 cm below its equilibrium position;
x = 32 cm = 0.32 m
g = 9.81 m/s²
We know that
k = mω²
where ω = 2π/T
So, k = m( 2π/T )²
we substitute
k = 0.12( 2π / 0.54 )²
k = 0.12 × (11.6355)²
k = 0.12 × 135.38486
k = 16.25 N/c
so Also,
kx²/2 = mgh
we solve for h
h = kx² / 2mg
we substitute
h = ( 16.25 × (0.32)²) / ( 2 × 0.12 × 9.81 )
h = 1.664 / 2.3544
h = 0.7068 m
Therefore, the wooden animal will go 0.7068 m above its equilibrium
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Answer:
F = 37.8 × 10^(6) N
Explanation:
The charges are 0.06 C and 0.07 C.
Thus;
Charge 1; q1 = 0.06 C
Charge 2; q2 = 0.07 C
Distance between them; r = 3 m
Formula for the force in between them is;
F = kq1•q2/r²
Where k is a constant = 9 × 10^(9) N.m²/C²
Thus;
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?
Answer:
d= 23.25 m
Explanation:
Assuming no other external forces acting on the disk, total mechanical energy must be conserved.Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:[tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]
When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:[tex]E_{f} = m*g*h (2)[/tex]
Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:[tex]h = d* sin 37 (3)[/tex]
Replacing (3) in (2):[tex]E_{f} = m*g* d * sin 37 (4)[/tex]
Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:[tex]v = \omega * R (5)[/tex]
For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².The rotational kinetic energy of a rotating rigid body can be written as follows:[tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]
Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:[tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]
Since (4) and (7) must be equal each other, we can solve for d as follows:[tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]
Replacing by the values, we finally get:[tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]
1.) Calculate the mass of a solid gold rectangular bar that has dimensions lwh = 4.30 cm ✕ 14.0 cm ✕ 27.0 cm. (The density of gold is 19.3 ✕ 103 kg/m3.)
kg
2.)A brass ring of diameter 10.00 cm at 17.3°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 17.3°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
(b) What if the aluminum rod were 10.06 cm in diameter?
Answer:
1) m = 0.3137 kg
2a)T_f = -181.7°C
2b) T_f = -1176.97°C
Explanation:
1) We are given;
Length; l = 4.30 cm = 0.043 m
Width; w = 14.0 cm = 0.014 m
height; h = 27.0 cm = 0.027 m
density of gold; ρ = 19.3 × 10³ kg/m³
Formula for the density is known as;
ρ = mass/volume
Thus;
m =ρV
m = 19.3 × 10³ × (lwh)
m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)
m = 0.3137 kg
2a) We are given;
Diameter of brass; L_br = 10 cm
Diameter of aluminum; L_al = 10.01 cm
Now, to some for change in temperature we will use the formula;
L_f = L_i + αL_i(Δt)
Where α is coefficient of expansion.
Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.
Thus;
L_f(brass) = L_f(aluminium)
From table attached, α_brass ≈ 19 × 10^(-6) /°C
Also, α_aluminium ≈ 24 × 10^(-6) /°C
Thus;
L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))
Similarly,
L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Since L_f(brass) = L_f(aluminium), then;
10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Rearranging, we have;
10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.01 = Δt(-50.24 × 10^(-6))
Δt = 0.01/(-50.24 × 10^(-6))
Δt ≈ -199°C
Thus, temperature at which the combination must be cooled to separate the two metals is;
T_f = T_i + Δt
T_f = 17.3 + (-199)
T_f = -181.7°C
2b) Diameter of aluminum is now;
L_al = 10.06 cm
Thus;
10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.06 = Δt(-50.24 × 10^(-6))
Δt = 0.06/(-50.24 × 10^(-6))
Δt = -1194.27°C
T_f = 17.3 + (-1194.27)
T_f = -1176.97°C
Which of the following quantities are unknown? initial separation of the particles final separation of the particles initial speed of the proton initial speed of the alpha particle final speed of the proton final speed of the alpha particle mass of the proton mass of the alpha particle charge of the proton charge of the alpha particle Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if A, C, and D are unknowns, enter ACD.
ALL of the quantities are unknown, because you haven't bothered to tell us anything that's known.
What order shows increasing frequency for gamma rays, microwaves, visible light, and X-rays?
gamma rays, X-rays, visible light, microwaves
microwaves, visible light, X-rays, gamma rays
visible light, gamma rays, microwaves, X-rays
X-rays, microwaves, gamma rays, visible light
Answer:
B. microwaves, visible light, X-rays, gamma rays
The order that shows increasing frequency for gamma rays, microwaves, visible light, and X-rays is:
microwaves, visible light, X-rays, gamma rays
What are electromagnetic waves?These are waves that can propagate (i.e travel) through space while transferring energy. They travel through space with the speed of light (i.e 3×10⁸ m/s)
Examples of electromagnetic waves includes
Gamma rayX-rayUltraviolet Light Infrared Radio wave MicrowaveFrom the examples given above, it should be noted that gamma ray has the shortest wavelength and the longest frequency.
With the above information in mind, we can conclude that the order that shows increasing frequency for gamma rays, microwaves, visible light, and X-rays is:
microwaves, visible light, X-rays, gamma rays
Learn more about electromagnetic waves:
https://brainly.com/question/8553652
A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.
Answer:
v = -17.94 cm
Explanation:
Given that,
The candle is placed at a distance of 50 cm, u = -50 cm
The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)
We need to find the image distance. We know that the lens formula is as follows:
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]
So, the image distance is equal to 17.94 cm.
The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light.
a. True
b. False
Answer:
False
Explanation:
No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
Which one of the following statements concerning the magnetic field inside (far from the surface) a long, current-carrying solenoid is true?
1) The magnetic field is zero.
2) The magnetic field is independent of the number of windings.
3) The magnetic field varies as 1/r as measured from the solenoid axis.
4) The magnetic field is independent of the current in the solenoid.
5) The magnetic field is non-zero and nearly uniform.
A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.
Answer: 84 %
Explanation:
Select the correct answer.
Each square dance begins with what?
A. Handshake
B. Dosado
C. Bow or curtsy
D. Promenade
Answer:
C
Explanation:
The men bow to the women and the women curtsy to the men.
(took on test and got it right)
10 POINTS! SPACE QUESTION!!
Answer; they are larger and made of rocky material
Describing a Wave
What does a wave carry?
Answer:
Waves carry energy from one place to another.
Explanation:
Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.
Momentum
Project: Egg Drop
Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your file(s) and are ready to upload your assignment, click the Add Files button below and select each file from your desktop or network folder. Upload each file separately.
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Answer:
get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it
Explanation:
Which investigation BEST measures the gravitational force on a
toy car?
A. rolling the car down a steep ramp and measuring time
B. using a spring scale and measuring the weight of the car
C. pushing the car and measuring how far it travels before it stops
D. throwing the car in the air and measuring how far it goes before coming
down
Answer:
B : using a spring scale and measuring the weight of the car
Explanation:
Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tuning fork and tune the piano to match, but her G4 tuning fork is missing! Instead, she plays the errant key along with her F4 tuning fork (which has a frequency of 349.2 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 76.7 Hz. Then, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz) and measures a beat frequency of 14.1 Hz.
What frequency is being played by the out-of-tune key ?
a. 363.3 Hz
b. 451.1 Hz
c. 33.9 Hz
d. 272.5 Hz
e. 425.9 Hz
Answer:
e. 425.9 Hz
Explanation:
The computation of the frequency is being played by the out-of-tune key is shown below;
Given that
Δf1 = x - 349.2 = 76.7.........(1)
Δf2 = 440 - x = 14.1......(2)
Now solve (1) and (2)
(440 - x) - x + 349.2 = 14.1 - 76.7
789.2 + (-2x) = -62.6
x = 425.9 Hz
Hence, the frequency is being played by the out-of-tune key is 425.9 Hz
Therefore the option e is correct
A toy car rolls down a ramp. Which force causes the car to move
Answer:
Gravity
Explanation:
Gravity pulls things down to earth and it is a force
The law of conservation of angular momentum states that if no external force acts on an object, then its angular momentum does not change. true or false
Answer:
the answer is false.
Explanation:
i took the test and it is false trust me!!!!!!!!!
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is 1.3 s.
What was the ball’s initial speed, in meters per second?
Answer:
[tex]u=14.48m/s[/tex]
Explanation:
From the question we are told that:
Height of window [tex]h=2m[/tex]
Height of window off the ground [tex]h_g=7.5m[/tex]
Time to fall and drop [tex]t=1.3s[/tex]
Generally the Newton's equation motion is mathematically given by
[tex]s=ut+\frac{1}{2}at^2[/tex]
Where
[tex]h=ut+\frac{1}{2}at^2[/tex]
[tex]2=u1.3-\frac{1}{2}*9.8*1.3^2[/tex]
[tex]2=u1.3-8.281[/tex]
[tex]u=7.91m/s^2[/tex]
Generally the Newton's equation motion is mathematically given by
[tex]2as=v^2-u^2[/tex]
Where
[tex]-2gh_g=v^2-u^2[/tex]
[tex]-2*9.8*7.5=(7.91)^2-u^2[/tex]
[tex]-147=62.5681-u^2[/tex]
[tex]u=\sqrt{209.5681}[/tex]
[tex]u=14.48m/s[/tex]
Therefore the ball’s initial speed
[tex]u=14.48m/s[/tex]
what happens when water at 4° celsius is heated further?
Answer:
please give me brainlist and follow
Explanation:
4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. ... Ice floats on top of lakes, preventing evaporation (and convection in the frozen layer), and lakes stay liquid underneath, allowing fish and other life to survive.
A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb
Answer:
C
Explanation:
C
What is the chemical formula for the molecule modeled?
Answer:
What is the chemical fórmula For the molecule modeled?
Explanation:
C6H12O2
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
mdjxjxjcjfkfjjdksklqlakzjxjxkkskakMmznxkxkdkd?
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I hope this helped!+*
what does loudness of a sound depend on?
Answer:
Amplitude
Explanation:
The loudness of a sound depends on the amplitude of vibration producing the sound
A 35.0 g bullet strikes a 50 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and the bullet fly off together at 8.6 m/s. What was the speed of the ballot before it struck the lumbar? Define the bullet and the wood as a system
Answer:
12294.31 m/s
Explanation:
Momentum = (mass)(velocity)
Momentum before = Momentum after
(momentum of bullet)+(momentum of block)=(momentum of bullet and block)
0.035v+50(0)=(0.035+50)(8.6)
0.035v=430.301
v=12294.3142857m/s