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3 years ago

Lake Baikal in Siberia has a maximum depth of 1642 m. What is the water

Physics

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Explanation:

Fluid gauge pressure is:

P = ρgh

where ρ is the fluid density and h is the depth of the fluid.

P = (1000 kg/m³) (9.8 m/s²) (1642 m)

P = 16,091,600 Pa

Rounded to four significant figures, the gauge pressure is 16.09 MPa.

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The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf

Galina-37 [17]

Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| 
E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J 

After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:

E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m 

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum

7 0

3 years ago

Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di

jeyben [28]

Answer :

(a) d = 0.25 m

(b) d = 0.5 m

Explanation :

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at $20^0\ C$ is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

$d=\dfrac{\lambda}{2}$ ........(1)

Velocity of sound wave is given by :

$v=f\times \lambda$

$d=\dfrac{v}{2f}$

$d=\dfrac{343\ m/s}{2\times 686\ Hz}$

$d=0.25\ m$

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

$d=n\lambda$ ( n = integers )

Let n = 1

So, $d=\dfrac{v}{f}$

$d=\dfrac{343\ m/s}{686\ Hz}$

$d=0.5\ m$

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0

3 years ago

Read 2 more answers

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$