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Viktor [21]
3 years ago
5

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

on in which such pressure fluctuations travel
Physics
1 answer:
DaniilM [7]3 years ago
4 0

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

        ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

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Loud of dust and gas in space is a(n) Blank Space __________.
madreJ [45]
A cloud in space that is composed of dust and gas is call a nebula. The word "nebula" is derived from the Latin word for "cloud," and nebulae in space are indeed large interstellar clouds made up of dust, hydrogen, helium and plasma.

6 0
3 years ago
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What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?
djyliett [7]
As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
</span>Therefore:
Internal energy = (5/2) nRT
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4 0
3 years ago
A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

6 0
3 years ago
Why does a third class lever cannot magnify force?​
maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

5 0
3 years ago
If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

1- We should cast off the bow and stern lines.

2-With the help of an oar or boat hook, keep the boat clear of the dock.

3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

7 0
3 years ago
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