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ExtremeBDS [4]
4 years ago
7

The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar m

ass of air is M=28.8×10−3kg/mol.
The questions are:
If the temperature at the surface is 28.00 ∘C and the temperature at a height of 260.0 m is 8.000 ∘C how long does it take sound to rise 260.0 m directly upward? [Hint: Use the equation v=sqrt(γRT/M) and integrate.]

At a height of 260.0 m, how far does sound travel horizontally in this same time interval?
Physics
2 answers:
Afina-wow [57]4 years ago
7 0

Answer: t = 0.878s

Explanation: Dear big brain

since your temperature decreases linearly, you can assume that your velocity should behave linearly too. Here this isn't exactly the case (cfr. formula). But there's another way to prevent endlessly boring calculus. Use the principle of interpolation. (x/v_surface + x/v_top)/2 = t.

This answer won't be exactly the same, but it's a quite good approx. You can use this eqation, hence you don't use large distances.

BARSIC [14]4 years ago
4 0

Answer:

0.771 s

Explanation:

v = sqrt(γRT/M) = dx/dt

t = int(1/v, x=0..260) = 260/sqrt(γRT/M)

In your case:

t = 260/sqrt((1.4*8.31*281)/(28.8*10^-3)) = 0.771 s

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4 years ago
The force f is given in terms of time t and displacement X by the equation F= Asin Bx +C sin dt. What is the dimension of D/B?​
denpristay [2]

Answer:

Here's what I think,

A_Trigonometric_ratio (whatever)

The general form of sin/cos/tan(angle).

So here as in the general form the "whatever" is always an "angle" input which (may sound a bit unfair but) is a dimensionless quantity.

As mentioned in your question,

F= Asin(Bt) + C cos (Dy)

Dimentions of these terms will be same (by homogenity of dimensions).

So what we gotta do here is make Bt and Dy dimensionless for this whole equation to work.

According to your question t is time,

So to make Bt dimensionless,

B must be equal to inverse of time i.e T^(-1)

Using the same process on Dy, gives inverse of length i.e L^(-1)

So,

B= M^(0)L^(0)T^(-1)

D = M^(0)L^(-1)T^(0)

The dimension of D/B then will be M^(0)L^(-1)T^(1)

6 0
3 years ago
Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and
Tems11 [23]

Answer:

<em>0.153 m/s</em>

<em></em>

Explanation:

The flowrate Q = 0.625 x 10-3 m^3-/s

The diameter of the nozzle d = 5.19 x 10^-3 m

the velocity V = ?

The cross-sectional area of the flow A = \pi d^{2}/4

==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2

From the continuity equation,

Q = AV

V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = <em>0.153 m/s</em>

4 0
4 years ago
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