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Fynjy0 [20]
4 years ago
11

What are longitudinal waves?

Physics
1 answer:
My name is Ann [436]4 years ago
3 0
Waves in which the displacement of the medium is in the same directions as, or the opposite direction to, the traveling wave's direction.

A wave in which the particle movement is parallel to the wave's direction.

I gave to definition, I hope you understand it. Have a nice day :)
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You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r
Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

d=ut-\dfrac{1}{2}gt^2

Putting all values

d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0
3 years ago
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0
3 years ago
Locomotion is the act of staying stationary. <br><br> True <br> False<br><br> 20 POINTS
Fittoniya [83]

Answer:

False

Explanation:

Locomotion is associated with movement.

The word is derieved from two Latin words,

Loco----place

Motio----to move.

8 0
3 years ago
A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai
Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air (\rho), measured in kilograms per cubic meter, is defined by following equation:

\rho = \frac{P\cdot M}{R_{u}\cdot T} (1)

Where:

P - Pressure, measured in kilopascals.

M - Molar mass, measured in kilomoles per kilogram.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 2026.5\,kPa, M = 28.965\,\frac{kg}{kmol}, R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} and T = 300\,K, then the density of air is:

\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}

\rho = 23.534\,\frac{kg}{m^{3}}

The mass of air stored in the vessel is derived from definition of density. That is:

m = \rho \cdot V (2)

Where m is the mass, measured in kilograms.

If we know that \rho = 23.534\,\frac{kg}{m^{3}} and V = 10\,m^{3}, then the mass of air stored in the vessel is:

m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})

m = 235.34\,kg

The mass of air stored in the vessel is 235.34 kilograms.

7 0
3 years ago
Imagine that you are approaching a black hole in a spacecraft. what would you see? what would happen to you?
Vladimir79 [104]
You will see some thing pitch black pulling you. you will feel an immense pull or attraction while going rapidly inside the black hole. 
3 0
3 years ago
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