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lisov135 [29]
3 years ago
15

An osprey's call is a distinct whistle at 2200 Hz. An osprey calls while diving at you, to drive you away from her nest. You hea

r the call at 2300 Hz. How fast is the osprey approaching?
I have tried this eqaution and the answer is still not correct..
vs= f+ - f-/f+ + f- (v) 2300-2200/2300+2200(343 m/s) =7.6 m/s. That is way too fast for a bird to fly? Can someone please help??
Physics
1 answer:
Elis [28]3 years ago
4 0

Answer:

14.9 m /s

Explanation:

n = n° x v /[ v -v (s) ]

n is apparent frequency , n° is source frequency,v is speed of sound and v(s)

speed of source.

2300 = 2200 x 343 / [343 -v(s)]

v (s ) = 14.9 m /s.

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Mary spots Bill approaching the dorm at a constant rate of 2.00 m/s on the walkway that passes directly beneath Mary's window, 1
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Answer:

Mary will have to wait for 63.2 seconds

Explanation:

Time required for the apple to drop from a height of 17.0 m above the ground to 1.65 m above the ground is given by the formula below:

t = √2h/g  where h is height through which the object falls, g is acceleration due to gravity

h = 17.0 - 1.65 = 15.35 m

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t = 1.77 s or approximately 1.8 s

Time taken for bill to get to the point below Mary's window is given below;

time taken = distance/velocity

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Therefore, Mary will have to wait for (65 - 1.8) s = 63.2 seconds

4 0
3 years ago
Junior slides across home plate during a baseball game. If he has a mass of 115 kg, and the coefficient of kinetic friction betw
tatyana61 [14]

Weight Force of Junior = m g = 115kg x 9.81 m/s^2 = 1128.15N then compute for the friction force


Friction Force= WF x (coefficient of kinetic friction) = 1128.15N x 0.35 =  394.8525N or 395N

 

But you can compute in a straightway:

Solution:

= 115 x 9.81 x 0.35

= 394.85

= 395 N

 

It will still give the same results.

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Which statement best describes the relationship between mass and gravitational attraction (pull)?
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Answer:

The first option, "The more mass an object has, the greater the gravitational pull. "

Explanation:

Newton's Law of Gravity states that F_G=\frac{Gm_1m_2}{r^{2}}, where G is the gravitational constant, m_1 and m_2 are the masses of the two objects, and <em>r</em> is the distance between the objects' centers. Because the objects' masses are in the fraction's numerator, they are directly proportional to F_G, and increasing the mass of one or both objects increases the gravitational pull.

Please have a great day! I hope this helps you understand the question!

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