The given statement "if the period of a pendulum is triple, then the frequency will also increase" is False.
If the period of a pendulum is tripled, then the frequency will decrease. The period and frequency of a pendulum are inversely proportional.
What is a pendulum?A pendulum is a weight suspended from a pivot that is free to swing back and forth due to the force of gravity. A classic example is a pendulum clock. The pendulum's back-and-forth motion is known as its oscillation. The time it takes for one complete oscillation, also known as one cycle, is known as the pendulum's period.
How are frequency and period related?Frequency and period are inversely proportional. The frequency of a wave or oscillation is the number of cycles it completes in one second, while the period is the amount of time it takes to complete one cycle. The frequency is calculated by dividing the number of cycles completed by the time taken to complete them.
The frequency and period are linked mathematically. Period = 1/frequency. The frequency and period of a pendulum are inversely proportional. A pendulum's frequency decreases as its period increases.
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A massive light hangs over the table in Jeremy's dining room. The light is supported by four strong chains which make an angle of 72° with the horizontal. The force in each chain is 36.4 N. Determine the mass of the light in kilograms (kg). Use the approximation g ≈ 10 m/s^2.
Given
A massive light hangs over the table in Jeremy's dining room. The light is supported by four strong chains which make an angle of 72° with the horizontal.
The force in each chain is F=36.4 N.
To find
The mass of the light in kg
Explanation
Let the mass of the light be m
The weight of the light acts downwards.
To balance thisi force the force on the string vertically upward is considered
In equillibrium
\(\begin{gathered} mg=4Fsin72^o \\ \Rightarrow10m=4\times36.4\times sin72^o \\ \Rightarrow m=13.84\text{ kg} \end{gathered}\)Conclusion
The mass of the light is 13.84 kg
what will most likely happen if a sound wave moves from the air through a solid
a. it will increase wavelength
b. it will decrease in speed
c. it will increase in speed
d. it will decrease in wavelength
Answer:
C. it will increase in speed
Explanation:
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PLSS HELPP
Are humans evolving? What will humans look like in 100,000 years in the future or longer?
Answer:
yes
Explanation:
A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitude of the final velocity when it strikes the ground?
The magnitude of the final velocity of the 2.00 kg ball, thrown upward at some unknown angle from the top of a 20.0 m high building, with an initial magnitude of velocity of 20.0 m/s is 44.3 m/s.
We are given Mass of the ball, m = 2.00 kg Initial velocity of the ball, u = 20.0 m/sHeight of the building, h = 20.0 mThe formula to find the final velocity of a body under free fall motion is given by:v² = u² + 2ghWhere, v = final velocity of the body u = initial velocity of the body g = acceleration due to gravity h = height through which the body falls Substituting the given values in the above formula, we get:v² = 20.0² + 2 × 9.81 × 20.0v² = 400 + 392.4v² = 792.4v = √792.4v = 28.13 m/sThe final velocity of the ball when it reaches the ground will be equal to the velocity with which it hit the ground. The ball is thrown upwards and comes back down, so we have to consider both the upward motion and the downward motion to find the final velocity.
The total time taken by the ball to reach the ground can be found by adding the time taken for the upward motion and the time taken for the downward motion 2t = (20.0/9.81) + (20.0/9.81)2t = 4.07t = 2.03 sThe final velocity of the ball when it reaches the ground is given by:v = u + gtWhere, v = final velocityu = initial velocityg = acceleration due to gravityt = total time taken by the body to reach the ground.Substituting the given values in the above formula, we get:v = 20.0 + 9.81 × 2.03v = 44.3 m/sTherefore, the magnitude of the final velocity of the 2.00 kg ball, thrown upward at some unknown angle from the top of a 20.0 m high building, with an initial magnitude of velocity of 20.0 m/s is 44.3 m/s.
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a wind tunnel can be used to model the effect wind has on an airplane. what is one limitation of this model?
A. Pieces of planes, such as wings, can be tested.
B. Highly skilled workers are necessary to run the wind tunnel.
C. The airflow in the tunnel can be varied to mimic atmospheric conditions
D. A wind tunnel provides valuable information about how planes fly
Answer: B. Highly skilled workers are necessary to run the wind tunnel
Explanation:
A pex
The limitation of this model is highly skilled workers are necessary to run the wind tunnel.
The answer is option B.
What are the restrictions of a wind tunnel?Tunnels open to the environment also are laid low with winds and climate. high running prices. The fan must continually boost up the float through the tunnel. Noisy operation.
What are the advantages of the usage of a wind tunnel to design a plane?They gauge elevate, drag, shockwaves, and other conditions that have an effect on planes and other devices that velocity through the wind. in addition, those tunnels can help engineers parent out how wind interacts with desk-bound items, along with buildings and bridges, and locate methods to make them stronger and safer.
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the center of the sun is powered by which type of energy?
Answer: Its nuclear fusion
Explanation:
Answer:
Pretty sure it's Radiation.
A student sits on a rotating stool holding two 1 kg objects. when his arms are extended horizontally, the objects are 0.8 m from the axis of rotation, and he rotates with angular speed of 0.69 rad/sec. the moment of inertia of the student plus the stool is 6 lg m? and is assumed to be constant. the student then pulls the objects horizontally to a radius 0.4 m from the rotation axis. calculate the final angular speed of the student answer in units of rad/s. calculate the change in kinetic energy of the sustem. answer in units of j.
The energy that an object has as a consequence of motion is known as kinetic energy. It is described as the effort required to propel a mass-determined body from rest to the indicated velocity.
What are a few instances of kinetic energy?An illustration of kinetic energy
Kinetic energy, which may be seen in the motion of an item or subatomic particle, is the energy in motion. Kinetic energy exists in every particle and moving object. Example of kinetic energy in action include a person walking, a baseball soaring through the air, a piece of food falling from a table, but a charge particles in an electric field.
What makes it kinetic energy?The energy from mass that is moving is known as kinetic energy. An element's kinetic energy is the power it possesses as a result of motion.
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A ship travels north at 15 km/h for 2 h. What is the ship's displacement in km?
The ship travels with a velocity of 15 km/h within 2 hours. The distance travelled is the velocity times the time taken that is 30 km to the north.
What is velocity ?Velocity is a physical quantity measuring the distance travelled by an object per unit time. Velocity is the rate of speed and it is a vector quantity having both magnitude and direction. The unit of velocity can be m/s, km/h, ft/s etc.
Velocity is the ratio of the change in distance to the change in time. The change in distance is sometimes called the displacement.
then velocity = displacement / time
Given that velocity = 15 km/h
time taken = 2 hours
then displacement = velocity × time
= 15 × 2 = 30 Km
Therefore, the displacement of the ship is 30 km to the north.
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Two long parallel wires carry currents of 20 a and 5. 0 a in opposite directions. the wires are separated by 0. 20 m. what is the magnetic field midway between the two wires? group of answer choices
The magnetic field will be given by the formula B=B1+B2.
Between two wires, the magnetic field has the formula B=B1+B2.
We must first calculate the magnetic field caused by each wire before adding the two vectors to find the net magnetic field halfway between the two wires. We must subtract the two vectors in order to find the total magnetic field because the electric current flows in the opposite direction via both cables.
The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is traveling through a magnetic field.
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If a bowling ball weighing 3,630 g is accelerating down the alley at a rate of 4 m/s2, then what is the net force applied to the bowling ball?
Answer:
\(\huge\boxed{\sf F = 14.52\ N}\)
Explanation:
Given data:Mass = m = 3,630 g = 3.63 kg
Acceleration = a = 4 m/s²
Required:Force = F = ?
Formula:F = ma
Solution:Put the given data in the above formula.
F = (3.63)(4)
F = 14.52 N\(\rule[225]{225}{2}\)
Question 4 Multiple Choice Worth 4 points)
(02.02 MC)
How do weathering and deposition differ?
Answer:
its A Weathering breaks down rocks; deposition leaves them in new places
Explanation:
During a demonstration of Newton’s laws of motion, a student used the setup shown in Figure 1. The student flicked the index card with a fingertip, and the coin fell straight down into a plastic cup as shown in Figure 2.
A: A falling body accelerates at a constant speed.
B: The motion of an object is constantly changing due to magnetic forces.
C: The force friction causes an object in motion to move faster.
D: An object remains at rest, or at a constant speed, an unbalanced force acts on it.
Answer:
D. An object remains at rest, or at a constant speed, an unbalanced force acts on it.The coin was at rest until the card was removed, so it tended to remain in the same location. Once the card was gone, the unbalanced force of gravity caused the coin to fall.
I hope this helped!+*♡
How does an annular eclipse differ from a total eclipse.
Answer:
Solar eclipses may be classified as either total, in which the Moon completely covers the Sun, or annular, in which the Moon obscures all but an outer ring of the Sun.
How much of a 1500. mg sample of potassium-42 is left after 48 hours if its half-life is 12.4hours?
Given data
*The initial quantity of potassium-42 is 1500 mg
*The given time is t = 48 hours
*The half-life of Potassium - 42 is T = 12.4 hours
The expression for the radioactive decay is given as
\(A(t)=A_0(\frac{1}{2})^{\frac{t}{T_{}_{}_{}}}\)Substitute the values in the above expression as
\(\begin{gathered} A(48)=1500(\frac{1}{2})^{\frac{48}{12.4}} \\ =102.59\text{ mg} \end{gathered}\)What does the half life for a radioactive material mean?
The electric field 0.385 m from a very long uniform line of charge is 810 N/C
How much charge is contained in a section of the line of length 2.50 cm?
The charge contained in a section of the line of length 2.50 cm is 8.87 × 10⁻¹⁰ C.
The formula for electric field intensity of a line charge is given by:E= λ/2πε₀rwhere,λ is the linear charge density of the line.ε₀ is the permittivity of free space.r is the perpendicular distance of the point from the line charge.
Electric field intensity, E = 810 N/CandDistance, r = 0.385 mUsing the above formula, we can find the value of linear charge density of the line.λ = 2πε₀Erλ = 2 × π × 8.85 × 10⁻¹² × 810 × 0.385λ = 3.55 × 10⁻⁸ C/mLength of the section of the line, L = 2.5 cm = 0.025 mWe need to find the charge present in a section of the line of length 2.50 cm.Since the linear charge density of the line is 3.55 × 10⁻⁸ C/m,Charge in a section of the line of length 0.025 m = λLq = λLq = 3.55 × 10⁻⁸ × 0.025q = 8.87 × 10⁻¹⁰ C
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when xenon absorbs 801 kj/mol of energy, it is excited into a higher-energy state in which the outermost electron has been promoted to the next available subshell. choose the right electron configuration for this excited xenon.
when xenon absorbs 801 kj/mol of energy, it is excited into a higher-energy state in which the outermost electron has been promoted to the next available subshell. The right electronic configuration for this excited xenon is [Kr] 4d¹⁰5s²5p⁴5d²
The energy required to completely remove the outermost electron from the excited xenon to form excited Xe+ is 369 kJ/mol, and here the energy absorb by the xenon is 801kj/mol which is twice the energy required to remove the 1st excited electron i.e. first ionisation energy, so 801kj/mol is the second ionisation energy of xenon.
The ground state electronic configuration is-
[Kr] 4d¹⁰5s²5p⁶
The excited electronic configuration of xenon is -
[Kr] 4d¹⁰5s²5p⁴5d².
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if his eyes have a far point of 2.0 m , what is the greatest distance he can stand from the mirror and still see his image clearly? express your answer using two significant figures.
The greatest distance the person can stand from the mirror and still see their image clearly is 2.0 meters.
The far point of a person's eyes refers to the maximum distance at which they can see objects clearly without the aid of corrective lenses. If the far point is given as 2.0 m, it means that the person can see objects clearly up to a distance of 2.0 m.
To determine the greatest distance the person can stand from the mirror and still see their image clearly, we need to consider the concept of the virtual image formed by a mirror. When we look into a mirror, our eyes perceive a virtual image as if it were formed behind the mirror.
In this case, the person's eyes can focus clearly up to a distance of 2.0 m. To see their image clearly in the mirror, the person needs the virtual image formed by the mirror to be within this range.
Since the virtual image formed by a mirror is the same distance behind the mirror as the object is in front of it, the person needs to stand at a distance from the mirror equal to the maximum distance they can focus clearly.
Therefore, the greatest distance the person can stand from the mirror and still see their image clearly is 2.0 meters.
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What acceleration would be achieved by a 5N thrust motor in a 0.30kg
Answer:
F=ma
5N=0.3a
a=5/0.3=16.66m/s²
PLEASE GIVE BRAINLIEST
Answer:
16.7
Explanation:
acceleration=force/mass
15. A ruby-throated hummingbird beats its wings at a rate of about 70 wing beats per
second.
What is the frequency in Hertz of the sound wave?
b. Assuming the sound wave moves with a velocity of 350 m/s, what is the
wavelength of the wave?
U
E
Answer:
a
Explanation:
A skier traveling downhill has this type of energy
Answer:
potential energy
Explanation:
one advantage of the hubble space telescope over ground based ones is that
Space telescopes have the advantage of being above the blurring effects of the Earth's atmosphere.
A surrounding is a layer of fuel or layers of gases that envelop a planet and is held in a region through the gravity of the planetary frame. A planet retains its surroundings whilst the gravity is exceptional and the temperature of the surroundings is low. A stellar ecosystem is the outer area of a celeb, which includes the layers above the opaque photosphere; stars of low temperature may have outer atmospheres containing compound molecules.
The atmosphere of Earth consists of nitrogen (78%), oxygen (21%), argon (0.9%), carbon dioxide (0.04%), and trace gases. most organisms use oxygen for breathing; lightning and bacteria perform nitrogen fixation to provide ammonia that is used to make nucleotides and amino acids; flora, algae, and cyanobacteria use carbon dioxide for photosynthesis.
The layered composition of the ecosystem minimizes the damaging results of daylight, ultraviolet radiation, solar wind, and cosmic rays to defend organisms from genetic harm. The modern composition of the atmosphere of the Earth is made from billions of years of biochemical modification of the paleo atmosphere by residing organisms.
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In which one of the following objects is the centre of gravity located outside the body?
regular square
equilateral triangle
a sphere
a ring
A circuit is arranged like in figure 4, what is the current in each resistor? V1=5V, V2=7V,V3=5V,V4=7V ans R1=30Ω,R2=50Ω,R3=30Ω,R4=60Ω and R5=25Ω. Be sure to show your work, especially your set-up steps (defining currents, picking loops, etc) Figure 4: V1=5V,V2=7V,V3=5V,V4=7V ans R1=30Ω,R2=50Ω,R3=30Ω, R4=60Ω and R5=25Ω
The approximate currents in each resistor are: In R1: I1 ≈ 0.077 A, In R2: I2 ≈ 0.186 A, In R3: I3 ≈ 0.263 A, In R4: I4 ≈ 0.098 A, In R5: I5 ≈ 0.165 A.
To solve for the current in each resistor in the given circuit, we can apply Kirchhoff's laws, specifically Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL).
First, let's label the currents in the circuit. We'll assume the currents flowing through R1, R2, R3, R4, and R5 are I1, I2, I3, I4, and I5, respectively.
Apply KVL to the outer loop:
Starting from the top left corner, move clockwise around the loop.
V1 - I1R1 - I4R4 - V4 = 0
Apply KVL to the inner loop on the left:
Starting from the bottom left corner, move clockwise around the loop.
V3 - I3R3 + I1R1 = 0
Apply KVL to the inner loop on the right:
Starting from the bottom right corner, move clockwise around the loop.
V2 - I2R2 - I4R4 = 0
At the junction where I1, I2, and I3 meet, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
I1 + I2 = I3
Apply KCL at the junction where I3 and I4 meet:
The current entering the junction is equal to the current leaving the junction.
I3 = I4 + I5
Now, let's substitute the given values into the equations and solve for the currents in each resistor:
From the outer loop equation:
V1 - I1R1 - I4R4 - V4 = 0
5 - 30I1 - 60I4 - 7 = 0
-30I1 - 60I4 = 2 (Equation 1)
From the left inner loop equation:
V3 - I3R3 + I1R1 = 0
5 - 30I3 + 30I1 = 0
30I1 - 30I3 = -5 (Equation 2)
From the right inner loop equation:
V2 - I2R2 - I4R4 = 0
7 - 50I2 - 60I4 = 0
-50I2 - 60I4 = -7 (Equation 3)
From the junction equation:
I1 + I2 = I3 (Equation 4)
From the junction equation:
I3 = I4 + I5 (Equation 5)
We now have a system of five equations (Equations 1-5) with five unknowns (I1, I2, I3, I4, I5). We can solve these equations simultaneously to find the currents.
Solving these equations, we find:
I1 ≈ 0.077 A
I2 ≈ 0.186 A
I3 ≈ 0.263 A
I4 ≈ 0.098 A
I5 ≈ 0.165 A
Therefore, the approximate currents in each resistor are:
In R1: I1 ≈ 0.077 A
In R2: I2 ≈ 0.186 A
In R3: I3 ≈ 0.263 A
In R4: I4 ≈ 0.098 A
In R5: I5 ≈ 0.165 A
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if an object weighs 700 n in air and 500 n in water, a. calculate the buoyant force it experiences in water b. what is the density of the object?
a. The object having weight 700N in air, experiences buoyancy force of 200N in water.
b. The density of the object is 3.5g/cm3
a. Archimedes Principle:
When a body is partially or fully immersed in a fluid, its weight appears to decrease and apparent decrease in its weight is equal to weight of the fluid displaced by the body.
The object weight in air is 700N and it's weight in water is 500N.
Buoyancy force = Apparent loss of weight
= 700N - 500N
= 200N
b. Density of the object:
Buoyancy force = Vdg ( where v is volume of the object, d is the density of liquid and g is gravity of earth)
200 = Vdg
200 = V × 1000 × 10 (density of water= 1000kg/m3)
V = 2/100
m/d = 2/100 (V = m/d ,where d is density of solid)
(mg = 700N given in question)
(m = 70)
d = 70× 100/2
d = 3500kg/m3
= 3.5g/cm3
hence density of the object is 3.5g/cm3
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How much energy is required to increase the temperature of a 10kg block of iron from 15^oC to 25^oC? The specific heat capacity of iron is 0.44J/kg^oC. Use the equation change in thermal energy=mass*specific heat capacity*temperature change
The amount of thermal energy required to increase the temperature of a 10kg block of iron from 15 °C to 25 °C is 44 J
How do I determine the thermal energy required?First, we shall list out the given parameters from the question. This is given below:
Mass of iron block (M) = 10 kilogramsInitial temperature (T₁) = 15 °CFinal temperature (T₂) = 25 °CChange in temperature (ΔT) = 25 - 15 = 10 °CSpecific heat capacity of iron (C) = 0.44 J/Kg°CThermal energy required (Q) =?The thermal energy required can be obtained by using the formular illustrated below:
Q = MCΔT
Q = 10 × 0.44 × 10
Q = 44 J
Thus, we can conclude that the thermal energy required is 44 J
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If a person is kicking there legs three times every four seconds what is the frequency.
Answer:
f = 0.75 Hz
Explanation:
It is given that, a person is kicking there legs three times every four seconds. We need to find his frequency. The number of times an event is occurring is called frequency of an object. The frequency of a person is given by :
\(f=\dfrac{n}{T}\\\\f=\dfrac{3}{4}\\\\f=0.75\ Hz\)
So, the frequency of the person is 0.75 Hz.
a 210 mw vertically polarized laser beam passes through a polarizing filter whose axis is 40 ∘ from horizontal.
The vertically polarized laser beam, after passing through the polarizing filter with a 40° angle from the horizontal, will have an intensity of 126 MW.
When a 210 MW vertically polarized laser beam passes through a polarizing filter, its intensity will be affected. In this case, the polarizing filter has its axis at a 40° angle from the horizontal.
The angle between the polarization direction of the laser beam and the axis of the polarizing filter determines the intensity of the transmitted light. When the angle is 0° or 180° (parallel or anti-parallel), the intensity of the transmitted light is maximum, while at 90° (perpendicular), the intensity is minimum.
In this scenario, the laser beam is vertically polarized (0° or 180°), and the polarizing filter's axis is at a 40° angle from the horizontal. Therefore, the angle between the polarization direction of the beam and the filter's axis is 40°.
The transmitted intensity can be calculated using Malus's law. According to Malus's law, the transmitted intensity (I) is given by I = I0 * cos²(θ), where I0 is the initial intensity (210 MW) and θ is the angle between the polarization direction and the filter's axis (40°).
Using the formula, the transmitted intensity can be calculated as follows: I = 210 MW * cos²(40°) = 210 MW * 0.6 = 126 MW.
Thus, the vertically polarized laser beam, after passing through the polarizing filter with a 40° angle from the horizontal, will have an intensity of 126 MW.
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Connected to a battery that offers a d.d.p. of 3.0Volts, we have the four lamps, all identical with resistances equal to 2.0 Ohms. The association appears in the figure. Note that the switch is open, so it doesn't pass current in the circuit (it's like an off switch). When the key closes, the lamps come on. Determining the Req of the association and the total current flowing through the circuit, we will find:
Answer:
Option D. 5Ω e 0.6 A
Explanation:
1. Determination of the equivalent resistance.
We'll begin by calculating the equivalent resistance of resistor B and resistor C. This can be obtained as follow:
Resistor B (R₆) = 2 Ω
Resistor C (R꜀) = 2 Ω
Equivalent Resistance (R₆꜀) =?
R₆꜀ = R₆ × R꜀ / R₆ + R꜀ (parallel connection)
R₆꜀ = 2 × 2 / 2 + 2
R₆꜀ = 4 / 4
R₆꜀ = 1 Ω
Finally, we shall determine the equivalent resistance of the circuit. This can be obtained as follow:
Resistor A (Rₐ) = 2 Ω
Equivalent resistance of resistor B and C (R₆꜀) = 1 Ω
Resistor D (Rₔ) = 2 Ω
Equivalent Resistance (R) =?
R = Rₐ + R₆꜀ + Rₔ (series connection)
R = 2 + 1 + 2
R = 5 Ω
Thus, the equivalent resistance of the circuit is 5 Ω
2. Determination of the total current in the circuit.
Potential difference (V) = 3 V
Equivalent resistance (R) = 5 Ω
Total current (I) =?
V = IR
3 = I × 5
Divide both side by 5
I = 3 / 5
I = 0.6 A
Thus, the total current in the circuit is 0.6 A
Summary:
Equivalent resistance = 5 Ω
Current = 0.6 A
Option D gives the current answer to the question.
The major concern involved in architectural acoustics is how A. indirect sound reflections change sound quality. B. direct sound reflections change sound quality. C. indirect sound reflections affect VAS. D. direct sound reflections affects VAS
A. Indirect sound reflections refer to the sound waves that bounce off surfaces in a room before reaching the listener.
The major concern involved in architectural acoustics is how indirect sound reflections change sound quality.
These reflections can affect the sound quality by altering the characteristics of the sound, such as its clarity, intelligibility, and reverberation.
Architectural acoustics aims to optimize the design and arrangement of spaces to control and manage these indirect sound reflections.
This involves techniques such as the strategic placement of sound-absorbing materials, the use of diffusers to scatter sound waves, and the control of room dimensions and shapes to minimize undesirable echoes and reverberation.
While direct sound reflections can also influence the sound quality, they are often less of a concern in architectural acoustics compared to indirect reflections.
Direct sound refers to the sound that reaches the listener without any significant interaction with the room's surfaces. However, the design of architectural spaces can still consider the control of direct reflections to improve sound clarity and intelligibility in specific scenarios.
Therefore, among the given options, A. indirect sound reflections changing sound quality is the primary concern in architectural acoustics.
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