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Westkost [7]
3 years ago
5

How do odd-shaped ceilings, decorative panels, draperies, and glass windows affect echo and noise?

Physics
1 answer:
vesna_86 [32]3 years ago
6 0
I think they decrease echo and reduce noise, they do this by either absorbing vibrations or by scattering the sound so that echoes arrive at different times rather than reverberating as a standing wave. An echo is a reflection of a sound that arrives at the listener with a delay after the direct sound. The delay is usually proportional to the distance of the reflecting surface from the source and the listener.
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I WILL MARK BRAINLIST PLEASE HELP!
Alenkasestr [34]

Answer:

Generally, when thermal energy is transferred to a material, the motion of its particles speeds up and its temperature increases. There are three methods of thermal energy transfer: conduction, convection, and radiation. ... Convection transfers thermal energy through the movement of fluids or gases in circulation cells.

Explanation:

4 0
3 years ago
Read 2 more answers
Which of the following graph is used for determining the instantaneous velocity from the slope?
hoa [83]

Answer:

B. x - t graph

Explanation:

A position-time (x-t) graph is a graph of the position of an object against (versus) time.

Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.

An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.

Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.

<u>For example;</u>

Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.

Solution.

S(t) = 4t^{2} + 2t + 10

Differentiating the equation, we have;

S(t) = 8t + 2

Substituting the value of "t" into the equation, we have;

S(5) = 8(5) + 2

S(5) = 40 + 2

S(5) = 42 m/s.

5 0
3 years ago
Read 2 more answers
Which of the following cannot be determined from the spectrum of a star?
Aneli [31]
I think the correct answer from the choices listed above would be the last option. It is the chemicals in the core of the star that cannot be determined from the spectrum of a star. Spectrum shows the different classification of the stars depending on their spectral characteristics. It usually involves the light, the wavelength and the distance.
4 0
3 years ago
The earth has a mass of 5.98 × 10^24 kg and the moon has a mass of 7.35 × 10^22 kg. The distance from the centre of the moon to
Leokris [45]

Answer:

F = 4.48N

Explanation:

In order to calculate the net gravitational force on the rocket, you take into account the formula for the gravitational force between two objects, which is given by:

F=G\frac{m_1m_2}{r^2}         (1)

G: Cavendish's constant = 6.674*10^-11 m^3kg^-1s^-2

r: distance between the objects

You have a rocket at the middle of the distance between Earth and Moon, then, you have opposite forces on the rocket.

If you assume the origin of a system of coordinates at the rocket position, with the Moon to the left and the Earth to the right, you have:

F=G\frac{M_em}{r_1^2}-G\frac{M_mm}{r_2^2}       (2)

Me: mass of the Earth = 5.98*10^24 kg

Mm: mass of the Moon = 7.35*10^22 kg

m: mass of the rocket = 1200kg

r1: distance from the rocket to the Earth = 3.0*10^8m

r: distance between rocket and Moon = 3.84*10^8m - 3.0*10^8m = 8.4*10^7m

You replace the values of the parameters in the equation (2):

F=Gm[\frac{M_e}{r_1^2}-\frac{M_m}{r_2^2}]\\\\F=(6.674*10^{-11}m^3kg^{-1}s^{-2})(1200kg)[\frac{5.98*10^{24}kg}{(3.0*10^8m)^2}-\frac{7.35*10^{22}kg}{(8.4*10^7m)^2}]\\\\F=4.48N

The net force exerted over the rocket is 4.48N

4 0
3 years ago
A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient
Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

W = \Delta KE

W = F*d

Where,

F= Force (Frictional at this case F_r = \mu N)

d= Distance

\Delta KE = \frac{1}{2} mv^2

Where,

m = mass

v = velocity

Equation both terms,

F*d = \frac{1}{2}mv^2

\mu mg *d = \frac{1}{2}mv^2

\mu g * d = \frac{1}{2}v^2

d = \frac{1}{2} \frac{v^2}{\mu g}

Replacing with our values we have that

d = \frac{1}{2} \frac{25^2}{0.65*9.8}

d = 49.05m

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

6 0
3 years ago
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