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4 years ago

Brainliest and 100 POINTS

Physics

2 answers:

anastassius [24]4 years ago

8 0

Reflection- serious thought or consideration

Absorption- the process or action by which one thing absorbs or is absorbed by another.

Diffusion- the spreading of something more widely.

Diffraction- the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.

Refraction- measurement of the focusing characteristics of an eye or eyes

Natasha_Volkova [10]4 years ago

6 0

Reflection is where the sound seems to bounce off the instrument and play back at its surroundings

Absorption is the opposite of reflection, where a physical object absorbs the sound rather than bounces it away.

Diffusion is an even spread of a sound across a given area

Diffraction is the bending of sound waves around barriers

Refraction is the bending of sound propagation trajectories

You might be interested in

Select the appropriate units for each property of a wave.

IceJOKER [234]

Intensity: Decibels

Amplitude: Meters

Frequency: Hertz

Explanation:

The Wave is not visible to eyes and they can easily propagate through vacuum. the average power travelling at a given period of time in a space is the intensity. Decibels is the measure of intensity. it is measured in the decibel scale. The wave's strength and the intensity gives the amplitude of wave. It is measured using meters.

The wave's amplitude and the energy has a direct proportionality. The number occurrence of wave cycles per second refers to the frequency of wave. it is measured in hertz. it is also measured as the number of cycles that occurs per second.

4 0

3 years ago

The velocity of a car increases from 10 km/h to 50 km/h in 5 seconds. What is its acceleration?

mr Goodwill [35]

Answer:

2.22m/s^2

Explanation:

3 0

3 years ago

Help me with this problem please? c:

deff fn [24]

30*3*60=5400
300*2=600
1500*3=4500
200*10*60=120000

the computer uses more energy.
(you can multiply all results by 60 to get the energy in J, but it's not necessary for the comparison)

5 0

3 years ago

A spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two

vampirchik [111]

Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

R₁ = 3.59 cm

R₂ = 7.22 cm

m₁ = m₂ = m

K₁ = K₂

We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² (I)

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² (II)

∵ K₁ = K₂

⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒ (ω₁ / ω₂) = 1.9079

8 0

3 years ago

A child on a sled has a combined mass of 97 kg. At the top of a 3.1 meter hill, the sled has a velocity of 8 m/s. Assuming there

Lemur [1.5K]

Answer:

Explanation:

This is an energy conservation problem in that the total energy available to a system s constant trhoughout the whole problem. That is, in equation form:

TE = PE + KE. This is a bit of a problem due to sig figs here, but I will keep with the rules for them all the way up til the end of the problem (I'll tell you when I veer away from the rule and why I did when I get there).

We first need to find the total energy available to the system and then use that value throughout the rest of the problem.

TE = PE + KE where

PE = mgh (mass times gravity times height of the object) and

KE = $\frac{1}{2}mv^2$ (one-half times the mass of the object times the square of the velocity of the object). We solve for PE first, rounding to the correct number of sig figs:

PE = 97(9.8)(3.1) and

PE = 2900 J (2 sig figs here since all the numbers given have 2 sig figs in them). Now for KE

$KE=\frac{1}{2}(97)(8.0)^2$ which, to 2 sig figs (notice I added a .0 to the 8), is

KE = 2200. This means that the total energy available to the sled throughout the whole trip is

TE = 2900 + 2200 so

TE = 5100 J

Now for the second part of the problem, the sled is at a different height so the PE is different, and we are asked to find the velocity at the top of this second hill. The total energy equation then is

5100 = PE + KE. Solving for PE first:

PE = (97)(9.8)(1.2) and, to 2 sig figs:

PE = 82. Now for the KE:

$KE=\frac{1}{2}(97)v^2$ . Plugging all of that into the total energy equation gives us:

$5100=82+\frac{1}{2}(97)v^2$ and isolating the v:

$v=\sqrt{\frac{2(5100-82)}{97} }$ and this is where I veer away from the sig figs. Doing the subtracting first, as we should, means that we would round to the hundreds place. Since the least significant place in the number 82 is the tens place and the least significant place in the 5100 is in the hundreds place, we round to the hundreds place. But I didn't keep to this rule because I'm not sure how adamant your physics teacher is about sig figs. I'll give you the answer that I would expect from my students (the right one) after I give you this one. I just simply subtracted 5100 - 82 and used the answer the calculator gave me which was 5018:

$v=\sqrt{\frac{2(5018)}{97} }$ gives us that the velocity at the top of the 1.2 meter hill is

10.2 m/s.

Doing this the right way, using the rules for sig figs properly:

$v=\sqrt{\frac{2(1500)}{97} }$ gives us, to 2 sig figs,

v = 1.0 × 10¹ m/s

5 0

3 years ago