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kiruha [24]
3 years ago
10

Two forces with magnitudes of 6 pounds and 18 pounds are applied to an object. The magnitude of the resultant is 13 pounds. Find

the measurement of the angle between the resultant vector and the vector of the 18 pound force to the nearest whole degree.
Physics
1 answer:
Nikolay [14]3 years ago
6 0

Answer: the angle between the resultant vector and the vector of the 18 pound is 28°

Explanation:

given that data in the question; as its interpreted in the diagram below;

from the cosine rule, we know that;

a² = b² + c² - 2bc  

so

(13)² = (6)² + (18)² - (2 × 6 × 18 ) cos∅

169 = 36 + 324 - 216cos∅

169 = 360 - 216cos∅

216cos∅ = 360 - 169

216cos∅ = 191

cos∅ = 0.8842

∅ = cos⁻¹ ( 0.8842 )

∅ = 27.8° ≈ 28°  {nearest whole number}

Therefore the angle between the resultant vector and the vector of the 18 pound is 28°

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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
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Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

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Explanation:

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The velocity as a function of time of a moving particle is given by v = α+ βt2 , where α and β are constants and t is time in s.
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