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Maru [420]
3 years ago
12

4. If the Kitty Hawk Flyer has 12 HP, do some research to find out how to express this power in the SI system and report the num

erical value in kilowatts.
Physics
2 answers:
lesya [120]3 years ago
8 0

Product of my research: 1 HP = 746 watts .

12 HP = (12 x 746 W) = 8,952 W

8,952 W = 8.952 kW
Alik [6]3 years ago
4 0

Explanation:

The Kitty Hawk Flyer has 12 HP. We need to express this power in the SI system and in kilowatts.

The SI unit of power is watts and HP is also one of the unit of power. The relationship between the watts and horse power is :

1 HP = 745.7 watts

To convert 12 HP to watts we will use the above relation. So,

12 HP = 8948.4 watts

Also, 1 kilowatts = 1000 watts

So, 8948.4 watts = 8.94 kilowatts

Hence, this is the required solution.

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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
_____________ is the study of movement in athletes. A. Sports biomechanics B. Posture C. Dynamics D. Anatomy
liubo4ka [24]

Answer:

I believe its A: Sports biomechanics.

3 0
3 years ago
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni
irinina [24]

The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

  • <em>length of the solenoid, L = 91 cm = 0.91 m</em>
  • <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
  • <em>number of turns of the solenoid, N = 1300 </em>
  • <em>current in the solenoid, I = 3.6 A</em>

The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

\mu_o is the permeability of frees space = 4π x 10⁻⁷ T.m/A

B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

Learn more here:brainly.com/question/17137684

7 0
2 years ago
Hi guys i need truly help!
lutik1710 [3]

Answer:

1) 327

2) 3

3) 109

------------

now I'm not 100% sure #4-6 are correct, but if 4 is right then 5 and 6 are, too.

4) 1074

5) 1074 + 109t

6) 2382

Explanation:

7 0
3 years ago
What one part of a circuit might cause your calculator to be off?
Tanzania [10]

Resistors tell me if im right

7 0
3 years ago
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