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Angelina_Jolie [31]
2 years ago
14

A mechanical high-speed bat is flying along a path perpendicular to a wall. it emits a sound with a frequency f0 . the night is

clear and the air is still. let the speed of sound waves in the still air be vs . the bat hears a sound wave reflected from the wall with a frequency fnew . from this information one can determine the speed vbat at which the bat is flying. install a detector at the wall. 0 bat f v what is the detected frequency? 1. f1 = vs + vbat vs f0 2. f1 = vs vs − vbat f0 3. f1 = vs vs + vbat f0 4. f1 = vs + vbat vs − vbat f0 5. f1 = 2 vs vs − vbat f0 6. f1 = vs − vbat vs f0 7. f1 = vs − vbat vs + vbat f0 8. f1 = f0 9. f1 = 2 vs vs + vbat f0
Physics
1 answer:
NARA [144]2 years ago
4 0
1. f1 = Vs + Vbat Vs fO
2. f1 = Vs Vs - Vbat fO
3. f1 = Vs Vs + Vbat fO
4. f1 = Vs + Vbat Vs - Vbat fO
5. f1 = 2Vs Vs - Vbat fO
6. f1 = Vs - Vbat Vs fO
7. f1 = Vs - Vbat Vs + Vbat fO
8. f1 = fO
9. f1 = 2Vs Vs + Vbat fO

The observed frequency is
f= (V+Vr) × fO/ (V + Vs)
The speed of the sound is V = Vs
The speed of the receiver is  Vr = O
The speed of the source
Vs = -Vbat where negative is the bat which is approaching the wall
Then the correct answer is 
f1 = (Vs / (Vs -Vbat)) × fO.
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Answer:0.2mm

Explanation:

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Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

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Mass of the skater is, m=68.5\ kg

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

F_c=\frac{mv^2}{R}

Now, plug in the given values of the known quantities and solve for centripetal force, F_c. This gives,

F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N

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550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
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Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

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P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

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