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docker41 [41]
3 years ago
15

2. A string with a length of 0.9m that is fixed at both ends

Physics
1 answer:
aliina [53]3 years ago
3 0

Answer:

2a.) Wavelength = 1.8 m

2b.) F = 66.67 Hz

3a.) Find the attached file

3b.) Wavelength = 0.6 m

Explanation:

Given that the

Length L = 0.9m

Wavelength (λ) = 2L/n

Where n = number of harmonic

If n = 1, then

Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m

b.)

 If waves travel at a speed of 120m/s on this string, what is the frequency

associated with the longest wave (first harmonic)?

Given that V = 120 m/s

V = Fλ

But λ = 2L, therefore,

F = V/2L

F = 120/1.8

F = 66.67 Hz

3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic

Using the same formula,

Wavelength (λ) = 2L/n

Where n = 3

Wavelength (λ) = 2 × 0.9/3

Wavelength (λ) = 0.6 m

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Explain why air masses do not mix
kati45 [8]

 

Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force that preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.

If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.

At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a "front". In the military jargon, a front is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.

3 0
2 years ago
Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done
sergiy2304 [10]
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
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3 0
2 years ago
Work is the transfer of _______ that occurs when a force makes an object move.
Vlad1618 [11]
I think it is energy
4 0
3 years ago
The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?
Degger [83]
There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.
7 0
3 years ago
Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc
TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0
3 years ago
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