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motikmotik
3 years ago
8

Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ

ency of 544 Hz. The other cyclist hears the frequency as 563 Hz. If the speed of sound in air is 344 m/s, what is the speed of the motorcycles
Physics
1 answer:
Galina-37 [17]3 years ago
5 0

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

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3 years ago
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A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,
iris [78.8K]

Answer:

The work done by gravity is 4.975 \: Joules

Explanation:

The data given in the question is :

Mass is 0.245 kg

Height from ground is 2.07 m

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

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Insert values from question

Work done = mass \times gravity \times (change \: in \: height)

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4 years ago
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

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\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

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v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

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Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

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Answer:

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