Answer:
False
Explanation:
i got it right
The statement is false. Because, the place where, it is easy to hear soft voices is acoustically good. Acoustically strong rooms facilitates the sound waves of even soft pitches.
What is Acoustics?Acoustics is the study of sound waves. Engineers strictly consider the concepts of acoustics to design the auditoriums and theater rooms to filter the sound effects in perfect pitches.
The frequency of a sound wave is called pitch. Thus lower frequency waves are lower pitch and they are termed as soft voices. Soft voices are hard to hear.
However, building where it is very easy to hear soft voices is acoustically good room. Therefore, the statement is false. The building is good in acoustics suitable for sound effects.
To find more on acoustics, refer here:
https://brainly.com/question/29842668
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I need help in question 9 plzzz
True is the correct answer
Name three items to remember when completing a job application.
1. ) It's good to remember contact information such as phone number or email.
2.) Your address, so if they need to mail anything to you, they can. This is also a necessary point of information on most job applications anyway.
3.) Past job experiences and ways to contact them. This is probably one of the most important things as it shows you've worked before and are willing to talk to past employers.
May I have brainliest please? :)
Answer:
the other answer is correct
Explanation:
i do flvs and this was on the career final exam
1. How much heat is produced if you heat .25 kg water from 25 degrees C to 100 degrees C. The specific heat of water is 4180 J/kg C.
Answer:
Q = 78375 [J]
Explanation:
To solve this problem we must use the following ideal equation for the thermal energy of an element or substance depending on the temperature change.
[tex]Q=m*C_{p}*(T_{final}-T_{initial})[/tex]
where:
Q = heat [J]
m = mass = 0.25 [kg]
Cp = specific heat = 4180 [J/kg*C]
Tinitial = 25[°C]
Tfinal = 100 [°C]
[tex]Q =0.25*4180*(100-25)\\Q = 78375 [J][/tex]
You get a job delivering water. You calculate how much work is done picking up each 20 L bottle of
water and raising it vertically 1 m. For every 100 bottles you deliver, you will use Select.... (g =
9.8 m/s2)
-196J
-2,000 J
-19,600 J
-196,000J
Answer:
The work done by picking up 100 20-L bottles and raising it vertically 1 meter is 19614 joules.
Explanation:
By the Work-Energy Theorem, the work needed to raise vertically 100 bottles of water is equal to the gravitational potential energy, units for work and energy are in joules:
[tex]\Delta W = \Delta U_{g}[/tex] (1)
Where:
[tex]\Delta W[/tex] - Work.
[tex]\Delta U_{g}[/tex] - Gravitational potential energy.
The work is equal to the following formula:
[tex]\Delta W = n\cdot \rho \cdot V \cdot g \cdot \Delta h[/tex] (2)
Where:
[tex]n[/tex] - Number of bottles, dimensionless.
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta h[/tex] - Vertical displacement, measured in meters.
If we know that [tex]n = 100[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]V = 0.02\,m^{3}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta h = 1\,m[/tex], then the work done is:
[tex]\Delta W = (100)\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot (0.02\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1\,m)[/tex]
[tex]\Delta W = 19614\,J[/tex]
The work done by picking up 100 20-L bottles and raising it vertically 1 meter is 19614 joules.
Describe two adaptations and explain how they help you survive.
Answer:
My feet and my fists
Explanation:
My feet help me walk
and my fists help me one punch people
Plzz Help easy quesion for 20 points Why does someone get burned if they touch an old filament bulb that has been on for a few hours? Plzz Help
Answer:
In reality, the filament gets so hot it in a real sense bubbles off molecules and electrons. Now and again this material gathers as a dull spot at the highest point of the bulb. Eventually, the filament falls apart, gets frail, and breaks, subsequently finishing the life of the light. Lights radiate light by siphoning an electric flow through a dainty tungsten fiber. The filament warms and emits light. Over the long haul, the filament oxidizes and turns out to be increasingly fragile, until it splits up and the bulb goes out. ... Tungsten picks up obstruction as it warms.
Hope this helped :)
Jack cannot run as fast as Paul. He decides to go at his own pace and not run with Paul. Which principle of fitness is he following?
Progression
Overload
Individuality
Reversibility
Answer:
individuality
Explanation:
since he cannot run with paul he decided to run by himself making his exercise individual
Answer: individuality
Explanation:
(you can ignore this its just the deffintion of individuality)
This is a crucial principle, the fundamental fact that everyone is different! Everyone responds to training in a different way. If you are walking or cycling with a friend, and doing exactly the same amount of training, don’t be concerned if one of you gets fitter faster than the other – this is what individualisation is all about.
It might be that one of you is having some pressure at work or difficulties at home, but wherever it is, it’s surprising what can affect your training. Some days your training can go really well and the next day, even though it was exactly the same length workout, it can be a nightmare. This is individualisation.
Find the maximum velocity possible for the person on the swing.
if someone could explain with work that would be very much appreciated since I wasn't there when this was discussed in class.
Answer:
The maximum possible speed is 4.85 m/s.
Explanation:
The Principle Of Conservation Of Mechanical Energy
In the absence of friction, the total mechanical energy is conserved. It means that :
[tex]E_m=U+K[/tex] is constant, being U the potential energy and K the kinetic energy
U=mgh
Where m is the mass of the object, g is the gravitational acceleration and h is the height from a fixed reference.
[tex]\displaystyle K=\frac{mv^2}{2}[/tex]
Where v is the speed.
When the person is at a maximum height of 2 meters, the speed is 0. Thus the mechanical energy is made only of potential energy.
Let's fix the reference for the height to the point where the person is at minimum height (1 m from the ground level). The maximum height with respect to this reference is h=2.2 m - 1 m = 1.2 m.
The potential energy is:
U = m*9.8*1.2
U = 11.76m
When the person is at the minimum (zero) height, the mechanical energy is made only of kinetic energy. Since the mechanical energy is conserved, then:
[tex]\displaystyle \frac{mv^2}{2}=11.76m[/tex]
Multiplying by 2 and simplifying by m:
[tex]v^2=2*11.76=23.52[/tex]
Solving for v:
[tex]v=\sqrt{23.52}=4.85[/tex]
[tex]\boxed{v=4.85}[/tex]
The maximum possible speed is 4.85 m/s.
AP physics here! Please help me answer only this question and then explain how you got your answer. After that, I will do the rest of the questions by myself using your explanation. Come here only if you know AP physics.
Answer:
displacement: 0
Distance: 30
Explanation:
lol, i asked my brother, he's also in 10th grade
Explain how the aperture geometry relates to the
diffraction pattern.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
how the aperture geometry relates to the
diffraction pattern:
Diffraction is the spreading out of waves as they pass through an aperture or around objects.it occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave.For every small aperture sizes,the vast majority of the wave is blocked.For large apertures the wave passes by or through the obstacle without any significant of diffraction.
in an aperture with width smaller than the wavelength,the wave transmitted through the aperture spreads all the way around the behave like a point sources of waves.
single slit diffraction pattern
The diffraction pattern made by waves passing through a slit of width [tex]\alpha[/tex] (larger than∫) can be understood by imagining a series of point sources all in phase along the width of the slit.The waves moving directly forward are all in phase,so they from a large central maximum.
if the waves travels at an angle Ф from the normal to the slit,then there is a path difference x between the waves production at the two end of the slit.
x=a sinΦ
The path difference between the top and middle waves is λ/2 then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.)at the angle so there is no resultant wave at this angle.Thus a minimum is the diffraction pattern is obtained at
λ=α sinθ
Now slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two section ,which is identical to the result of pairing off matching sources in the bottom two sections.in this case we obtained from the minimum.
λ/2=α/4 sinθ
we can divided the slit aperture into six equal sections and pair off sources in the top two divisions and then the bottom two,to give destructive interference for every matched pair.The minimum of intensity are obtained at angles
nλ = α sinθ
where n is an integer (1,2,......), but not n=0.There is a maximum of intensity in the center of the pattern. This process only gives the position of the minima,does not work for positions of the maxima,and so does not give the intensities of the maxima.
A star has ______________________________________________; it appears to be moving across the sky as a night progresses.
Answer:
Apparent Motion
Explanation:
Apparent Motion of a star is a scientific term that describes the phenomenon that is observed as a result of the earth's rotation which is based on the area or where the observer is situated, and then the point at which the star is found comparable to the rotation axis of the earth.
This often makes the start to appear like it's moving across the sky.
Hence, A star has APPARENT MOTION; it appears to be moving across the sky as the night progresses.
Define one joule heat.
Answer:
The joule is the standard unit of energy in electronics and general scientific applications. One joule is defined as the amount of energy exerted when a force of one newton is applied over a displacement of one meter. One joule is the equivalent of one watt of power radiated or dissipated for one second.
Hope it helps ^^
Please help me out
please show solvings
Answer:
5) The approximate height is about 80 m. Choice (c)
6) The stone takes about 3 seconds to rise to its maximum height
Explanation:
Free Fall Motion
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the gravitational acceleration, which value is [tex]g = 9.8 m/s^2[/tex].
The distance traveled by a dropped object is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
Question 5
Given the stone reaches the ground in t=4 seconds, the height of the tower is:
[tex]\displaystyle y=\frac{9.8*4^2}{2}=78.4\ m[/tex]
The approximate height is about 80 m. Choice (c)
Question 6
Vertical Motion
The vertical motion of an object is controlled by the force of gravity. This means that there is a non-zero net force acting on the object that makes it accelerate downwards.
If the object is thrown upwards at speed vo, its speed at time t is:
[tex]v_f=v_o-g.t[/tex]
The stone reaches its maximum height when the final speed is zero, thus:
[tex]v_o-g.t=0[/tex]
Solving for t:
[tex]\displaystyle t=\frac{v_o}{g}[/tex]
The stone is thrown vertically upwards with vo=30 m/s, thus:
[tex]\displaystyle t=\frac{30}{9.8}[/tex]
[tex]t=3.06\ s[/tex]
[tex]t\approx 3\ s[/tex]
The stone takes about 3 seconds to rise to its maximum height
A 45kg sled is being pulled from camp by 5 dogs each capable of exerting 25N force on the sled. If the sled starts from rest and the frozen ground exerts 15N of friction, how far will the sled be from the camp after 7s?
Answer:
58.8m
Explanation:
My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and stay another 60 minutes. We can ______________________ to see about how long we have to remain at the surface to have enough no stop time. (choose all that apply)
Answer:
1) Periodically check the no stop or NDL time on their computers
2) The dive computer planning mode can be used if available
3) Make use of a dive planning app
4) Check data from the RDP table or an eRDPML
Explanation:
The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop
The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning
The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.
A swing held up by someone at one side has the potential to swing when released.
True
False
HELP PLEASE BOYLES LAW
Answer:
3 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2 L
Initial pressure (P₁) = 0.75 atm
Final pressure (P₂) = 0.5 atm
Final volume (V₂) =?
Using the Boyle's law equation, the new volume (i.e final volume) of the Ne gas can be obtained as:
Initial volume (V₁) = 2 L
Initial pressure (P₁) = 0.75 atm
Final pressure (P₂) = 0.5 atm
Final volume (V₂) =?
P₁V₁ = P₂V₂
0.75 × 2 = 0.5 × V₂
1.5 = 0.5 × V₂
Divide both side by 0.5
V₂ = 1.5 / 0.5
V₂ = 3 L
Thus, the new volume of the Ne gas is 3 L
Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest
Answer:
The rank of the frequencies from largest to smallest is
The largest frequency of oscillation is given by the string in option D
The second largest frequency of oscillation is given by the string in option B
The third largest frequency of oscillation is given by the string in option A
The smallest frequency of oscillation is given by the string in option C
Explanation:
The given parameters are;
The mass per unit length of all string, m/L = Constant
The tension of all the string, T = Constant
The frequency of oscillation, f, of a string is given as follows;
[tex]f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}[/tex]
Where;
T = The tension in the string
m = The mass of the string
L = The length of the string
n = The number of overtones
[tex]Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K[/tex]
For the string in option A, the length, L = 27 cm, n = 3 we have;
[tex]f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K[/tex]
For the string in option B, the length, L = 30 cm, n = 4 we have;
[tex]f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K[/tex]
For the string in option C, the length, L = 30 cm, n = 3 we have;
[tex]f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K[/tex]
For the string in option D, the length, L = 24 cm, n = 4 we have;
[tex]f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K[/tex]
Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;
1 ) [tex]f_D[/tex] 2) [tex]f_B[/tex] 3) [tex]f_A[/tex] 4) [tex]f_C[/tex]
The order of the frequencies is [tex]f_D>f_B>f_A>f_C[/tex]
Standing waves:The frequency of the standing wave in a string tied at both ends is given by:
[tex]f=\frac{nv}{2L}[/tex]
where n is the mode of frequency
v is the velocity of the wave
and L is the length of the string.
Now the velocity of a wave in a string tied at both ends is given by
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where T is the tension and μ is the mass per unit length.
Since T and μ are the same for all the strings, velocity [tex]v[/tex] will be the same for all.
Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.
[tex]f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}[/tex]
Hence, [tex]f_D>f_B>f_A>f_C[/tex]
Learn more about standing waves:
https://brainly.com/question/1698005?referrer=searchResults
An airplane is flying at an elevation of 6 miles on a flight path that will take it directly over a radar tracking station. Let S represent the distance between the radar station and the plane. If S is decreasing at a rate of 400 mph when S is 10 miles, what is the velocity of the plane
Answer:
[tex]-500\ \text{mph}[/tex]
Explanation:
h = Height at which the plane is flying = 6 miles
S = Distance between plane and radar = 10 miles
[tex]\dfrac{dS}{dt}[/tex] = Rate at which S is decreasing = -400 mph
Distance between S and and the elevation of the plane
[tex]b=\sqrt{S^2-h^2}=\sqrt{10^2-6^2}\\\Rightarrow b=8[/tex]
From Pythagoras theorem we get
[tex]S^2=b^2+h^2[/tex]
Differentiating with respect to time we get
[tex]2S\dfrac{dS}{dt}=2b\dfrac{db}{dt}+2h\dfrac{dh}{dt}\\\Rightarrow S\dfrac{dS}{dt}=b\dfrac{db}{dt}+h\dfrac{dh}{dt}\\\Rightarrow 10\times -400=8\times \dfrac{db}{dt}+0\\\Rightarrow \dfrac{db}{dt}=\dfrac{10\times -400}{8}\\\Rightarrow \dfrac{db}{dt}=-500\ \text{mph}[/tex]
Velocity of the plane is [tex]-500\ \text{mph}[/tex].
If the distance between the radar station and the plane is decreasing at the given rate, the velocity of the plane is -500mph
Given the data in the question;
Elevation; [tex]x = 6miles[/tex]Distance between the radar station and the plane; [tex]S = 10miles[/tex]Since "S" is decreasing at a rate of 400 mph; [tex]\frac{ds}{dt} = -400mph[/tex]As illustrated in the diagram below, we determine the value of "y"
Using Pythagorean theorem:
[tex]S^2 = x^2 + y^2[/tex]------------Let this be Equation 1
We substitute in our value
[tex]y = \sqrt{(10miles)^2- (6miles)^2} \\\\y = \sqrt{100mi^2- 36mi^2}\\\\y = \sqrt{64mi^2}\\\\y = 8miles[/tex]
Now, we determine velocity of the plane i.e the change in distance in horizontal direction ([tex]\frac{dy}{dt}[/tex])
Lets differentiate Equation 1 with respect to time t
[tex]2S\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]------ Let this be Equation 2
Since, the plane is not landing, [tex]\frac{dx}{dt} =0[/tex]
We substitute our values into Equation 2 and find [tex]\frac{dy}{dt}[/tex]
[tex][2*10mi*(-400mph) ] = [2*6mi*0] + [2*8mi * \frac{dy}{dt}]\\\\-8000m^2ph = 0 + 16miles * \frac{dy}{dt}\\\\\frac{dy}{dt} = \frac{-8000m^2ph}{16miles}\\\\\frac{dy}{dt} = -500mph[/tex]
Therefore, if the distance between the radar station and the plane is decreasing at the given rate, the velocity of the plane is -500mph
Learn more: https://brainly.com/question/18187424
When going from a fast speed to a slow speed how is light bent?
1.away from the normal line
2.toward the normal line
Answer:
Toward the normal line
Explanation:
I had the same question on a quiz
A girl runs at a speed of 3.9 m/s off a high dive and hit the water 1.8 s later.
a. How high was the diving board?
b. How far horizontally was she from the board when she hit the water?
c. If she had just dropped off the board, would her time to drop to the water been longer, shorter or the same?
Answer:
(a) the height of the diving board is 22.896 m
(b) the horizontal distance traveled by the girl is 7.02 m
(c) if she had just drop off the board, her time to drop to the water would have been longer.
Explanation:
Given;
initial speed of the girl, u = 3.9 m/s
time to hit the water, t = 1.8 s
(a) the height of the diving board is calculated as;
h = ut + ¹/₂gt²
h = (3.9 x 1.8) + ¹/₂ x 9.8 x 1.8²
h = 7.02 + 15.876
h = 22.896 m
(b) the horizontal distance traveled by the girl is calculated as;
X = ut
X = 3.9 x 1.8
X = 7.02 m
(c) if she just drop off the board, then the initial speed will be zero;
h = ut + ¹/₂gt²
h = 0 + ¹/₂gt²
2h = gt²
[tex]t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2 \ \times\ 22.896 }{9.8} }\\\\t = 2.16 \ s[/tex]
Thus, if she had just dropped off the board, her time to drop to the water would have been longer.
What would be the net force on an object if the object's acceleration is 0?
Answer:
Explain. yes it is possible because when the net force is zero and acceleration is zero. ... no, Neglecting air friction, the only force acting on this object is the force of gravity downward. There is no force of equal magnitude as the force of gravity acting in the opposite direction.
A 747 requires a takeoff speed of 270 meters per second if it is to get off the
ground. Calculate the acceleration required for the 747 that begins from rest
to takeoff on a 1000 meter long runway.
Answer:
a = 36.45[m/s²]
Explanation:
We can calculate the acceleration value by means of the following expression of kinematics.
[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]
where:
vf = final velocity = 270 [m/s]
Vo = initial velocity = 0 (begins from rest)
a = acceleration [m/s]²
x = distance required by the plane = 1000 [m]
[tex](270)^{2} =0+2*a*1000\\72900=2000*a\\a=36.45 [m/s^{2} ][/tex]
can someone please explain some calculations on vernier calliper
Answer:
Hey mate here's your answer ⤵️
Vernier caliper least counts formula is calculated by dividing the smallest reading of the main scale with the total number of divisions of the vernier scale.LC of vernier caliper is the difference between one smallest reading of the main scale and one smallest reading of vernier scale which is 0.1 mm 0r 0.01 cm
Hope it was helpfulllKate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the mountains in the distance. This is an example of
Answer: motion parallax
Explanation:
Motion parallax refers to a form of depth perception whereby objects that are closer to an individual appears to move at a faster speed than the objects that are far.
Therefore, Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the mountains in the distance is an example of motion parallax.
what should we do to be a computer engineer ?
Answer:
to be an eginere u would have to go to college and study hard
Explanation:
different between group IA and IIA elements
Answer:
Group IA elements have only one valency electron while Group IIA have two valency electrons.
Group IA elements have cations with higher charge density hence polarizing anions easier resulting into covalent character while Group IIA elements have cations with lower charge density hence difficulty in distorting anions resulting into a ionic character. This is due to difference in cationic radii and charges
2. __________is the pressure exerted by the layer of air around the Earth
Answer:
Atmospheric Pressure
Explanation:
Imagine you've recently insulated your loft. Give a reason why your heating bill may still increase.
Answer:
follow me for the answer
Why does a third class lever cannot magnify force?
Explanation:
The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.