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katrin [286]
3 years ago
11

Bernoulli's principle is responsible for most of the lift produced by an airplane wing.

Physics
1 answer:
Rina8888 [55]3 years ago
5 0

Answer:

huiiiiuuu beautiful...

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12. Where is the Moon? If you see it, describe its location and appearance. ​
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What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!
Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

 657

×  32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

   ↳            21024 g/mL

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A 110-kg tackler moving at a speed of 3.5 m/s
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3 years ago
Where did we use rotational and irrotational flow​
WARRIOR [948]

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

  • If the fluid is rotating along its axis with the streamline flow of its particles,then this type of flow is rotational flow.
  • Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
  • In majority, if the flow of fluid is viscid then it is rotational.
  • Fluid in a rotating cylinder is an example of rotating flow.
3 0
3 years ago
A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

8 0
3 years ago
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