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notsponge [240]
2 years ago
11

On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s

. the speed of sound in air is v≈(331+0.60t)m/s, where t is the temperature in ∘c.
Physics
1 answer:
xenn [34]2 years ago
6 0
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
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Answer:

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Explanation:

Given that;

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f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

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6 0
2 years ago
A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H
aniked [119]

Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

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  • t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996  light-years

This distance is<em> very slightly less than 4 light-years. </em>

4 0
2 years ago
A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be
USPshnik [31]
<span>f2 = f0/4 The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius. The expression for the force of gravity is F = G*m1*m2/r^2 where F = Force G = Gravitational constant m1,m2 = masses involved r = distance between center of masses. Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after. f0 = G*m1*m2/r^2 f2 = G*m1*m2/(2r)^2 f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2) The Gm m1, and m2 terms cancel, so f2/f0 = (1/(2r)^2) / (1/r^2) f2/f0 = (1/4r^2) / (1/r^2) And the r^2 terms cancel, so f2/f0 = (1/4) / (1/1) f2/f0 = (1/4) / 1 f2/f0 = 1/4 f2 = f0*1/4 f2 = f0/4 So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
6 0
1 year ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

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=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
1 year ago
A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0
2 years ago
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