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jasenka [17]
2 years ago
13

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

d vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 45 m/s and accelerating at 1.1 m/s2. A train is approaching you from the opposite direction on the other track at 46 m/s, accelerating at 1.6 m/s2, and sounds its whistle, which has a frequency of 459 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz How fast is it changing? (Round your answer to two decimal places.) Hz/s
Physics
1 answer:
Alexus [3.1K]2 years ago
7 0

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

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Imagine you want to get 1 kcal of energy from a cow. How much energy would the cow need to get from plants? Why?
ZanzabumX [31]
1000 kcal because you only get 10% of the energy of the thing you eat
7 0
2 years ago
. 30
schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0
2 years ago
A projectile of mass m is fired horizontally with an initial speed of v0​ from a height of h above a flat, desert surface. Negle
Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0
2 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0
2 years ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
2 years ago
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