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lesya [120]
3 years ago
13

A train traveling 80.0 kph is blowing its horn as it approaches a railroad crossing. The horn has a frequency of 300.0 Hz. Assum

e the speed of sound is 331.5 m/s. What is the observed frequency of the horn?
395 Hz

281 Hz

322 Hz
Physics
2 answers:
marusya05 [52]3 years ago
8 0
The answer to your question is 322 hz
h
Alenkinab [10]3 years ago
8 0

Answer:

322 Hz

Explanation:

v = speed of train approaching the railroad crossing = 80 km/h = 80 x 1000/3600 m/s = 22.22 m/s

V = speed of sound of the horn of train = 331.5 m/s

f = actual frequency of the sound from the horn = 300.0 Hz

f' = observed frequency of the horn

Using Doppler's effect, observed frequency is given as

f' = V f/(V - v)

inserting the values

f' = (331.5) (300.0)/(331.5 - 22.22)

f' = 322 Hz

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Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N
nadezda [96]
First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say F_{1}, and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
F_{1} = 7i, where i = Unit vector in the x-direction.

2) Take the second force, say F_{2}, and assume that it is making an angle \alpha with the first force F_{1}.

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

F_{2} = (15*cos \alpha)i + (15*sin \alpha )j

Now from simple vector addition, we know that,
F_{R} = F_{1} + F_{2} --- (A)

Where F_{R} = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find F_{1}+F_{2} first!
F_{1}+F_{2} =  7i+(15*cos \alpha)i + (15*sin \alpha )j

=> F_{1}+F_{2} =  (7+15*cos \alpha)i + (15*sin \alpha )j

Now the magnitude of F_{1}+F_{2} is,
| F_{1}+F_{2}| = \sqrt{ (7+ 15*cos \alpha)^{2} +  (15*sin \alpha )^{2}}

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}

Since (sin \alpha)^{2} + (cos \alpha)^{2} = 1, therefore,

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Since  | F_{1}+F_{2}| = |F_{R}|, and the magnitude of the resultant force is 20N, therefore,

 |F_{R}| = | F_{1}+F_{2}|
20 = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Take square on both sides,
400 = 49 + 225 + 210*(cos \alpha)
(cos \alpha) =  \frac{3}{5}

\alpha = 53.13^{o}

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0
3 years ago
Object a travels in the +x-direction before hitting a stationary object
Leto [7]
The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:
 
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
 
Now, we know, tan</span>Ф = \frac{y}{x}

⇒tanФ = \frac{6.2 × 10^4 }{5.7 × 10^4}

⇒tanФ = 1.088

⇒ Ф = tan^{-1} 1.088 
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8 0
3 years ago
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Answer:

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What provided evidence for the revolution of earth around the sun?.
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Explanation:

there are many things. I am not sure what your teacher told you.

one of the first strange observations leading to this conclusion was the "loop-di-loop" in the track of Mars. because of the different orbits of Earth and Mars, sometimes one (particularly Earth as the faster one) swings around the sun in relation to the other, and with that first moves away and suddenly moves closer again, making the track of Mars appear on Earth as if it would make a loop in the sky.

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