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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??

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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A car travels from rest to 85 m/s and covers 150 meters. How LONG did<br> that take? *

kherson [118]

Answer:

51.76

Explanation:

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3 years ago

A motobike's tire rotates with a constant angular speed of 62.8 rad/s. The radius of a tire is 30cm. Assuming that no slipping o

Dmitriy789 [7]

Answer:

12.6 m/s

Explanation:

We know that the linear speed v = rω where r = radius and ω = angular acceleration. Given that ω = 62.8 rad/s and r = 20 cm from center = 0.2 cm

So, v = rω

= 0.2 m × 62.8 rad/s

= 12.56 m/s

≅12.6 m/s

7 0

3 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

4 years ago

abody starts from rest and accelerate uniformly at 8m/s/s.calculate the distance travelled by the body in 10s?

vagabundo [1.1K]

Explanation:

using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.

S=ut+½gt² ("ut" term will cancel because u=0).

=> S= ½gt²

=>S = ½×8×10²

=>S = 4×100

=>S = 400m .

Therefore, the distance traveled by the body in 10s is 400m.

hope this helps you.

6 0

2 years ago

LABORATORY TITLE:

mezya [45]

Answer:

PAPER CLIPS ON NOSE OF A PAPER AIRPLANE

Purpose: To determine if the number of paperclips on the nose of a paper airplane affects the velocity and speed, measured in meters per seconds.

Make a Hypothesis Based on the Learning Thus Far: If the number of paperclips on the nose of a paper airplane increases, then the speed will _increase______ (increase, decrease, stay the same) in a __linear_______ (linear, exponential, logarithmic) mathematical relationship, and the velocity will (increase, decrease, stay the same) in a __exponential____ (linear, exponential, logarithmic) mathematical relationship. (Fill in the appropriate words for your hypothesis.)

Pictures: Insert at least 3 pictures of yourself conducting the experiment into this lab report. At least 2 pictures must show your face as you conduct the investigation. You may need to ask someone to help take these photos.

Explanation:

5 0

3 years ago