Answer:
we measure sound intensity in Decibels.
Answer:
DecibelsExplanation:
We measure sound power or sound pressure in decibels.
They were named in honour of Alexander Graham Bell,( the inventor of both the telephone and the audiometer).
The flight path of a jet aircraft as it takes off is defined by the parmetric equations x=1.25 t2 and y=0.03 t3, where t is the time after take-off, measured in seconds, and x and y are given in meters. If the plane starts to level off at t = 40 s, determine at this instant:a. The horizontal distance it is from the airport b. Its altitudec. Its speed.
Answer:
A. We are given t= 40s, x=1.25t²
So The horizontal distance it is from the airport will be = 1.25(40)²= 2000meters
B. y= 0.03t³
So y= 0.03(40)³= 1920m altitude
C. To find speed
We say
dx/dt=Vx= 2.5t= 100
dy/dy = 0.09t= 144
So speed= √100²+144²= 175.31m/s
g Explain completely, but in the fewest words possible, why a ball thrown horizontally follows the path that it does
Answer:
Explanation:
A ball thrown horizontally is a type of projectile which moves in a trajectory path. It moves with respect to its momentum and gravitational pull of the earth.
The motion of the thrown ball can be described in two components, horizontal and vertical. In horizontal component of motion of the thrown ball, gravitational force is zero. While in the case of vertical component, the gravitational pull causes the path followed.
The force of gravity pulls the ball towards the surface of the earth gradually, thus forming a trajectory path.
The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use E_SCE = 0.241V.
Pt(s)│H2 (1.00 atm)│RNH2 (0.100M), RNH3 + (0.0500M)║SCE
Answer: Kb = 1.89 × 10⁻⁶
Explanation:
R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)
Ecell = - ( 0.0592 / n ) log Kₐ
Where, Ecell = 0.731 - 0.241 = 0.490 V
Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ
Therefore, Kₐ = 5.30 × 10⁻⁹
Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is [tex]f_{objective } = 43.0 \ cm[/tex]
The focal length of A is [tex]f_{eye} = 10.4 \ cm[/tex]
The theoretical angular magnification is mathematically represented as
[tex]m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}[/tex]
[tex]m = \frac{f_{objective }}{f_{eye}} = 4.175[/tex]
Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range
A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle is 420 Hz.
1. What is the wavelength of the sound waves in front of the locomotive? Use 344 m/s for the speed of sound in air.
2. What is the wavelength of the sound waves behind the locomotive? Use 344 m/s for the speed of sound in air.
3. What is the frequency of the sound heard by a stationary listener in front of the locomotive? Use 344 m/s for the speed of sound in air.
4. What is the frequency of the sound heard by a stationary listener behind the locomotive? Use 344 m/s for the speed of sound in air.
Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz
Explanation:
Give the following :
Velocity of train (Vt) = 26m/s
Frequency of sound (Fs) = 420Hz
Speed of sound (Vs) = 344m/s
1) wavelength = (Vs - Vt) / Fs
Wavelength = (344 - 26) / 420 = 318/420 = 0.757m
11) Wavelength = (Vs + Vt) / Fs
Wavelength = (344 + 26) / 420 = 370/420 = 0.881m
111) According to the doppler effect :
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source
Vs = - ve (train moving towards listener)
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = [(344 + 0) / (344 - 26)] * 400
Fl = (344 / 318) * 400 = 432.70Hz
1V) Vs = + ve (train moving away listener)
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = [(344 + 0) / (344 + 26)] * 400
Fl = (344 / 370) * 400 = 371.89Hz
(Application) You are asked to design a multiple lens system for the scenarios below. The lenses youhave available to you are:1. converging (f =7.5 cm)2. converging (f =15.0 cm)3. diverging (f = -15.0 cm)Think about the following scenarios:A) Lens 1 makes a real image i1, of which lens 2 makes a real image i2.B) Lens 1 makes a real image i1, of which diverging lens 2 makes a virtual image i2.C) Converging lens 1 makes a virtual image i1, of which converging lens 2 makes a virtual image i2.D) Converging lens 1 makes a virtual image i1, of which lens 2 makes a real image i2.For each scenario, decide:1) What type of lens is lens 1? Lens 2? Why? 2) Where should the object be placed with respect to the focal point to make the indicated type of image?3) Make a schematic drawing for Scenario D only. If you want extra practice, try to makeschematic drawings of the other three scenarios.
Answer:
1) A) Lens converging, B) lens 1 converging
2) A and B object is at a greater distance than its focal
3)
Explanation:
For this exercise we will give some more important characteristics of the lenses.
we use the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length and p and qlas the distance to the object and image, respectively.
We must also use how the lenses are combined
1 / = 1 / f₁ + 1 / f₂
where feq is the equivalent focal length and it is assumed that the two lenses are in contact.
By applying the equation of the constructor you get some characteristics of the image
Converging lens
* Object farther than focal length real and inverted image
* Oject after focal virtual image and right
Diverging lens
* for all virtual image distance and right
with this information let's analyze the different configurations
1) let's do an analysis for each scenario
A) since lens 1 and lens 2 create real image they must be converging lenses
B) Lens 1 must be convergent and lens 2 can be of both types
C) Lens 1 can be convergent with the object that is at a shorter distance than at focal point (p <f₁). The same analysis diverges for lens 2, the object that is the image of the other lens is shorter than its focal distance p₂ <f₂
D) the object is at a shorter distance than the focal length p₁ <f₁, the object for this lens is at a greater distance than its focal point p₂> f₂
2) let's analyze for each scenario
A) for each lens the object is at a greater distance than its focal length p₁> f₁ and p₂> f₂
B) for lens 1 p₁> f₁
for lens 2
if it is convergent p₂ <f₂
and if it is divergent any distance is possible
C) lens 1 p₁ <f₁ and for lens 2 p₂ <f₂
D) lens 1 p₁ <f₁ and for lens 2 p₂> f₂
3) see attached
The distance with subscript 1 is measured with respect to line1 and the distance with subscript 2 is measured with respect to the lens with subscript 2
According to Hookes Law, in order for a spring to apply a restorative force, what must be true about the displacement of the spring? a. The displacement must be proportional to the spring constant b. The displacement must be towards the spring, to compress it and cause it to force back c. The displacement must be measured away from a starting rest position in order to have a force from the spring d. The displacement must be away from the spring, to stretch it and cause it to pull back
Answer:
Its not A..
Explanation:
That was my choice and it was incorrect
The displacement must be towards the spring, to compress it and cause it to force back.
The displacement must be away from the spring, to stretch it and cause it to pull back.
What is the spring constant?The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.
The spring's restoring force attempts to return it to its equilibrium position; as a result, the displacement must be away from the spring in order to stretch it and make it a drawback, and it must be toward the spring in order to compress it and force it back.
Thus, the correct answers are option B and option D.
Learn more about spring constant,
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Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle at 16.2 m/s.
Answer:
The maximum height to which water could be squirted with the hose is 13.380 meters.
Explanation:
A line of current of a fluid can be explained sufficiently by Bernoulli's Theorem. In this case, the system can be simplified due to neglectance of changes in absolute pressure. Water is squirted with an initial speed and reaches its maximum height, where final speed is zero. That is to say:
[tex]\frac{v_{1}^{2}}{2\cdot g} +z_{1} = \frac{v_{2}^{2}}{2\cdot g} +z_{2}[/tex]
Where:
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final height of water, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of water, measured in meters per second.
If [tex]z_{1} = 0\,m[/tex], [tex]v_{1} = 16.2\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:
[tex]z_{2} = \frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} +z_{1}[/tex]
[tex]z_{2} = \frac{\left(16.2\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +0\,m[/tex]
[tex]z_{2} = 13.380\,m[/tex]
The maximum height to which water could be squirted with the hose is 13.380 meters.