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4 years ago

A(n) __________ is a fast-flowing river of air at the boundary between the troposphere and stratosphere.

1 answer:

7 0

<span>This is the jet stream. This air current flows from west to east along a typically-meandering path that is often used as a predictor of weather conditions. These currents are typically formed from the fact that the equator is warmer than the poles, and this allows the different-density air masses to develop an eastward-moving component.</span>

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Carey and Marcy are performing a lab to determine the speed of sound in air with the apparatus that was demonstrated in the vide

andre [41]

Answer:

the frequency is the fundamental and distance is L = ¼ λ

Explanation:

This problem is a phenomenon of resonance between the frequency of the tuning fork and the tube with one end open and the other end closed, in this case at the closed end you have a node and the open end a belly, so the wavelength is the basis is

λ = 4 L

In this case L = 19.4 cm = 0.194 m

let's use the relationship between wave speed and wavelength frequency and

v = λ f

where the frequency is f = 440 Hz

v = 4 L f

let's calculate

v = 4 0.194 440

v = 341.44 m / s

so the frequency is the fundamental and distance is

L = ¼ λ

5 0

4 years ago

The velocity of a particle traveling in a straight line is given by v = (6t - 3t²) m/s, where t is in seconds. If s = 0 when t =

qaws [65]

Answer

given,

v = (6 t - 3 t²) m/s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx = (6t - 3 t^2)\int dt$

x = 3 t² - t³

Position of the particle at t= 3 s

x = 3 x 3² - 3³

x = 0 m

now, particle’s deceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}(6t - 3 t^2)$

a = 6 - 6 t

at t= 3 s

a = 6 - 6 x 3

a = -12 m/s²

distance traveled by the particle

x = 3 t² - t³

at t = 0 x = 0

t = 1 s , x = 3 (1)² - 1³ = 2 m

t = 2 s , x = 3(2)² - 2³ = 4 m

t = 3 s , x = 0 m

total distance traveled by the particle

D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s

D = 2 + 4 + 2 = 8 m

average speed of the particle

$v_{avg} = \dfrac{distance}{time}$

$v_{avg} = \dfrac{8}{3}$

$v_{avg} =2.67\ m/s$

8 0

3 years ago

What do astronomers use to calculate the age of the universe

kykrilka [37]

The age of the oldest stars ,

Patterns of background radiation ,

How fast distant galaxies are moving away from us.

6 0

3 years ago

Read 2 more answers

-What do you think happened to make the Moon look the way it does?

goldenfox [79]

Answer: The physics of evolution had made the moon like it is today....Please watch this video from you tube about the evolution of the moon.

Explanation:

https://youtu.be/UIKmSQqp8wY

5 0

3 years ago

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

3 0

4 years ago