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3 years ago

7

What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round w

indow

1 answer:

ruslelena [56]3 years ago

6 0

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

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A 12,000kg. railroad car is traveling at +2m/s when it

ivann1987 [24]

<u>Answer:</u>

The final velocity of the two railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

8 0

3 years ago

Help ME QUICK PLEASE QUESTION 2 AND 4 PLEASE 58 POINTS

monitta

Answer:

For number 4: A vector pointing to the right with a magnitude of 2.0

Explanation:

Very simple- just subtract 6-2

I am not sure how to do #2- sorry!

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3 years ago

Two stereo speakers mounted 4.52 m apart on a wall emit identical in-phase sound waves. You are standing at the opposite wall of

Delicious77 [7]

The answer is option c) 0.9 m

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3 years ago

There’s an image here . Please help out !

ra1l [238]

Answer:

magnet 4

because opposite direction i.e north and south will attract each other

6 0

3 years ago

Read 2 more answers

g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

6 0

3 years ago