Answer: Some of the ways to control noise pollution are as follows: (1) Control at Receiver's End (2) Suppression of Noise at Source (3) Acoustic Zoning (4) Sound Insulation at Construction Stages (5) Planting of Trees (6) Legislative Measures.
Answer:
You have to:
a) Improve your insulation.
b) Install a fence
c) Use modern Acoustic wall panels
d) Plant trees
e) Reduce electronic volumes,e.t.c.
Explanation:
okay.
Is there a way for us to control motion
Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
Suppose an experiment is designed to test the durability of batteries in different conditions. All of the batteries tested are double-A (AA) Brand X. All sets of batteries are preconditioned in different environmental conditions for exactly 168 hours (1 week).
Set 1: 0°C (freezing point of water)
Set 2: 24°C (approximately room temperature)
Set 3: 37°C (approximately body temperature)
The batteries are then continuously used to power identical mechanical drummer toys. As long as the toy keeps drumming the battery is considered functional. The drumming time for each toy is measured as an indication of battery durability. In this experiment, which condition is not controlled?
A.) temperature
B.) brand of batteries
C.) test for durability
D.) type of battery (battery size)
Answer:
I assume its c. Since its talking about testing.
Explanation:
Answer:
The answer is test of durability
Explanation:
Which value would complete the last cell?
(1 point)
3.0
100.0
25.0
4.0
Answer:
4.0
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as
Force (F) = mass (m) x acceleration (a)
F = ma
With the above formula, we can obtain th acceleration of the body as follow:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
F = ma
20 = 5 x a
Divide both side by 5
a = 20/5
a = 4 m/s²
Therefore, the value that will complete the last cell in the question above is 4.
A block is attached to the end of a spring. The block is then displaced from its equilibrium position and released. Subsequently, the block moves back and forth on a frictionless surface without any losses due to friction. Which one of the following statements concerning the total mechanical energy of the block-spring system this situation is true?
1. The total mechanical energy is dependent on the maximum displacement during the motion.
2. The total mechanical energy is at its maximum when the block is at its equilibrium position
3. The total mechanical energy is constant as the block moves back and forth.
4. The total mechanical energy is only dependent on the spring constant and the mass of the block.
Answer:
The correct option is;
3. The total mechanical energy is constant as the block moves back and forth
Explanation:
The total mechanical energy is the sum of the potential and kinetic energies of the system
For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant
For a black and spring system, we have total mechanical energy, E = 1/2×K×A².
Where;
K = Constant
A = The amplitude of motion
Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.
what are some factors that affect the frequency of sound
Answer:
1. direction of propagation of sound
2.medium through which sound trsnsmitted
what is a hypothesis reffered to as after being verified by a large number or independent experiments
Answer:
The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.
The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points
Answer:
c
Explanation:
The marginal cost curve image has been attached from which we can clearly, indicate that
ATC = average total cost
AFC = average fixed cost
AVC = average variable cost.
From the graph we can indicate that the marginal cost curve
(c) Lies above the AVC curve when the AVC curve slopes downward.
Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.
Answer:
a)0.51°
b)1.47×10^-4m
Explanation:
a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.
bsin(θ)= mλ.........................(*)
Where b= width of the slit
The distance on the screen from Central angle can be expressed as
Sin(θ)= y/d............. (**)
d and y is the horizontal distance between slit and screen
If we input eqn(**) into equation (*) we have
y= mλd/b................(z)
In order to find angle (θ) we have
(θ)= sin-(1.99×10^-2)/2.25
= 0.51°
Therefore, angle of diffraction θ of the second minimum is 0.51°
b)to find the width of the sloth using eqn(z) by substitute the values, we have
b= (2)(649×10^-9)(2.25)/1.99×10^-2
b= 1.47×10^-4m
Therefore, the width of the slit is 1.47×10^-4m
you walk 6 block east, 2 blocks north, 3 blocks west and then 2 blocks north. the total distance you travel is blocks
Answer:
The answerI travel 13 blocksA car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Give your answer in correct SI units rounded to 0 decimal places.
Answer:
The distance the car travels is 115500 m in S.I units
Explanation:
Distance d = vt where v = speed of the car and t = time taken to travel
Now v = 99 km/h. We now convert it to S.I units. So
v = 99 km/h = 99 × 1000 m/(1 × 3600 s)
v = 99000 m/3600 s
v = 27.5 m/s
The speed of the car is 27.5 m/s in S.I units
We now convert the time t = 70 minutes to seconds by multiplying it by 60.
So, t = 70 min = 70 × 60 s = 4200 s
The time taken to travel is 4200 s in S.I units
Now the distance, d = vt
d = 27.5 m/s × 4200 s
d = 115500 m
So, the distance the car travels is 115500 m in S.I units
PLS HELP ME Define Derived Quantities ?
Derived Quantities
Explanation: Those physical quantities which are derived from fundamental quantities are called derived quantities and their units are called derived units. e.g., velocity, acceleration, force, work etc.
Answer:
These are quantities calculated from two or more measurements
Explanation:
They can't me measured directly.
They can only be computed.
They are calculated in PHYSICAL SCIENCE.
hope it helps.
A car is moving on straight highway with a speed of 108 km/h.
Answer:
5.3333 sec
Explanation:
initial speed: u = 108km/hr or 30 m/s
final speed: v = 0m/s
distance travelled: s = 80m
time the car took to stop: = t sec
[tex]v^{2} - u^{2}[/tex] = 2as,
a = ([tex]v^{2} - u^{2}[/tex])/2s
a = (0-900)/160
a = -5.625 [tex]ms^{-2}[/tex]
v = u + at,
t = (v - u)/a
t= (0 - 30)/(-5.625)
t = 5.3333 sec
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer? The nail exerts an equal force on the hammer in the same direction. The nail exerts a much smaller force on the hammer in the opposite direction. The nail exerts an equal force on the hammer in the opposite direction. The nail exerts a much smaller force on the hammer in the same direction.
Answer:
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.
Hello, I am BrotherEye
Answer:
Answers are
1. "The nail exerts an equal force on the hammer in the opposite direction."
2. "500 N"
3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."
4. "When mass moves closer to the point of rotation, rotational inertia decreases."
5. "The skater spins slower because his rotational inertia has increased."
Explanation:
Forensic toxicologist analyze and identify drugs that are confiscated from criminals
True
False
If the spring constant is 10 N/m and the spring is stretched 1 m, what is the Force?
Answer:
10N
Explanation:
Applying the Hooke law:
F = kx
F: Force
k: stiffness coefficient
x: stretched distance
F = 10N/m x 1m = 10N
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
Describe the motion of water waves.
Answer:
Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.
You and your family are going on a trip in Europe. Calculate the speed in the following scenario. State how you calculated your answer and include correct units. A car travels 240 kilometers in 3 hours; What is the speed of the car during that time?
Answer:
22.2 m/s or 80 km/h
Explanation:
Given that
Distance travelled by the car, d = 240 km
Time taken by the car, t = 3 hours.
Speed of the car, v = ? m/s
for easy calculations, we will be converting the units to meters and seconds respectively.
240 km to meters would be
240 * 1000 m = 240000 m
3 hrs to seconds would be
3 * 60 mins * 60 seconds = 10800 s
now, we have our distance and time to be
d = 240000 m
t = 10800 s
speed is defined as the ratio of distance with respect to time taken, effectively,
Speed = distance/time
speed, v= 240000 / 10800
v = 22.2 m/s
therefore, the speed of the car during the time is 22.2 m/s, or if the speed is needed in km/h, we can convert it
22.2 * 3600/1000 =
80 km/h
Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. When astronomers make such a statement, how can they know what was happening inside galaxies way back then
Answer:
First, as you may know, the light travels at a given velocity.
In vaccum, this velocity is c = 3x10^8 m/s.
And we know that:
distance = velocity*time
Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.
This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)
Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.
The astronomers could know what was happening inside galaxies way back then by the fact that;
they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find
Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.
This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.
The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.
Read more at; https://brainly.com/question/15995216
observe the virual skateboarder coming down the hill and over the ramp describe how each of newton’s laws of motion can be observed in this action you can choose the dry wet or muddy conditions or some combination of these
Answer:
first part the skater goes down a constant slope ramp, initially he has Newton's second law
pply Newton's third law, the normal is the reaction to the support of the body on the surface
the ramp shoots off. axis becomes zero and therefore with Newton's first law its speed
Explanation:
It is the description of this movement let's write Newton's laws.
* The first law that a body goes at constant speed or zero if the sum of the external forces is zero
* the second law is F = m a
* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.
Let's apply these laws to our case
In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.
The expression for this is
Wₓ - fr = ma
W sin θ - μ W cos θ = m a
W = mg
g (sin θ - μ cos) = a
the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy
This is Newton's second law
On the Y axis, which is perpendicular to the ramp we have
N- [tex]W_{y}[/tex] = 0
If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.
In the second part when he is on the ramp.
In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x
In this case the analysis is similar to the previous one
Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.
At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law
Answer:look at explanations
Explanation:
a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force was applied?
Answer:1m
Explanation:
2n=0.4m
5n=?
5n×0.4/2n=1m
an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil
Answer:
1) The density of the object = 2500 kg/m³
2) The density of the oil = 1250 kg/m³
Explanation:
1) The information relating to the question are;
The mass of the object in air = 0.250 kgf
The mass of the object in water = 0.150 kgf
The mass of the object in the oil = 0.125 kgf
By Archimedes's principle, we have;
The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced
The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf
∴ The weight of the water displaced = 0.1 kgf
Given that the object is completely immersed in the water, we have;
The volume of the water displaced = The volume of the object
The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)
The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³
The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³
The density of the object = 2500 kg/m³
2) Whereby the mass of the object in the oil = 0.125 kgf
The upthrust of the oil = The weight of the oil displaced
The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil
The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf
The weight of the oil displaced = The upthrust of the oil
Given that the volume of the oil displaced = The volume of the oil, we have;
The volume of the oil displaced = 0.0001 m³
The mass of the 0.0001 m³ = 0.125 kg
Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.
The density of the oil = 1250 kg/m³.
The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
Answer:
Positive
Explanation:
In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive
A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Answer:
The answer is The acceleration is double its original value.
Explanation:
It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.Hope this helps....
Have a nice day!!!!
Answer:
The acceleration is half of its original value
Explanation:
element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an atomic mass of 29.018 and a relative abundance of 17.67%. calculate the atomic mass of element X. show your work
Answer:
27.34 (no unit)
Explanation:
26.975*82.33%+29.018*17.67%
=27.34
How to find average acceleration only using displacement and time?
Answer:
Average acceleration = Displacement / (time)^2
Explanation:
The unit for acceleration is
[tex]m {s}^{ - 2} [/tex]
Displacement = m
Time = s
Hence the units of displacement and time should be manipulated to get the unit of acceleration.
You can't. You can only find average velocity.
But if you also know that initial velocity is zero ... the object started from rest ... then
Avg acceleration =
2 x displacement / time-squared
(b) A cylinder of cross-sectional area 0.65m2 and
height 0.32m has a mass of 2. Ikg. If there is a
cavity inside, find the volume of the cavity.
(Density of cylinder = 11.053 kg/m^3)
Answer:
The volume of the cavity is 0.013m^3
Explanation:
To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:
Step one:
Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.
Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]
Step two:
From the volume of the cylinder, we can get the radius of the cylinder.
[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]
Step three:
From the cross-sectional area, we can obtain the radius of the cavity.
Let the radius of the cavity be = r, while the radius of the cylinder be = R
CSA of cavity =
[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]
Step Four:
calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]
Recall that the cavity has the same height as the original cylinder
[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]
lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they:________.
Answer: React with the skin
Explanation:
lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they React with the skin
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 23.5 m/s at an angle of 39.5∘ above the horizontal. When the softball leaves her hand, it is 11.5 m above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Answer:
66.86m
Explanation:
Velocity of ball thrown, u = 23.5 m/s
Initial height of the ball above the water, H = 11.5 m
Angle of projection, θ = 39.5°
Vertical components of veloclty = usinθ
Horizontal components of veloclty = ucosθ
The soft ball hits the water after time 't'
Considering the second equation of motion
S = ut + 1/2at^2........ 1
But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:
H = ut +/- 1/2gt^2
H = - 11.5m
U = usinθ
θ = 39.5°
a = -g = -9.8m/s^2
- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2
-11.5m = 23.5(0.6360)t - 4.9t^2
-11.5m = 14.946t - 4.9t^2
4.9t^2 -14.946t-11.5m = 0
Since the ball drifted horizontally
D = (Ucosθ)t
Where θ = 39.5°
U = 23.5m/s t=
Alternatively,
horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s
now how long does it take the ball to raise to a peak and fall to the water.
vertical component of velocity = 23.5 sin 39.5º = 14.947m/s
time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec
peak reached above cliff top is
h = ½gt² = ½(9.8)(1.5252)²
= ½×22.797
= 11.3985m
now the ball has to fall 11.3985+ 11.5 = 22.8985m
time to fall from that height is
t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec
add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869
now haw far does the ball travel horizontally in that time
d = vt = 18.1331 ×3.6869= 66.856m
= 66.86m
