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DerKrebs [107]
3 years ago
9

What's a summary or abstract about cat eye syndrome?

Physics
1 answer:
Tanya [424]3 years ago
5 0

Answer:

This is the answer of your question.

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Sports managers have the potential to earn more than a million dollars per year.<br> True<br> false
andriy [413]

Answer:

true

Explanation:

hope this helped!

4 0
3 years ago
Sound waves need a medium to move through in order to keep moving. Through which of the following would sound travel the slowest
Troyanec [42]

Answer:A air

Explanation:Of the three mediums (gas, liquid, and solid) sound waves travel the slowest through gases, faster through liquids, and fastest through solids. Temperature also affects the speed of sound.

Hope this helps you out ツ

8 0
3 years ago
Read 2 more answers
Acceleration is equal to the initial velocity minus the final velocity, then divided by time.true or false
tatiyna

Almost true but not quite.

That would give you the negative of the actual acceleration.

It should be the other way around:

(final v) minus (initial v), then divide by time.

5 0
3 years ago
PE=?, m=.6kg, g=10m/s2, h=35m<br><br> PLS HELP I NEED THIS DONE
Karo-lina-s [1.5K]
210J

PE is mgh in this context.
7 0
4 years ago
An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
4 years ago
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