Physics, 08.10.2019 23:00 jnsoccerboy3121

The helicopter in the drawing is moving horizontally to the right at a constant velocity. the weight of the helicopter is w=52400 n. the lift force l generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) what is the magnitude of the lift force? (b) determine the magnitude of the air resistance r that opposes the motion.

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Answer from: Gheaton1889

We have that for the Question it can be said that the magnitude of the lift force and magnitude of the air resistance R

F=56128NR=20114N

From the question we are told

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=52400 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance R that opposes the motion.

Generally the equation for the lift force is mathematically given as

$F=\frac{w}{cos\theta}\\\\F=\frac{52400}{cos21}$

F=56128N

Generally the equation for the magnitude of air resistance is mathematically given as

$R=Lsin\theta\\\\R=56128N*sin 21$

R=20114N

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The helicopter in the drawing is moving horizontally to the right at a constant velocity. the weight

Answer from: leannamat2106

Explanation:

It is given that,

The weight of the helicopter, W = 52400 N

The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. The attached figure shows the free body diagram of the helicopter.

(a) Let L is the magnitude of the lift force.

The net force acting in the vertical direction,

$L\ cos\theta-W=0$

$L=\dfrac{W}{cos\ \theta}$

$L=\dfrac{52400}{cos(21)}$

L = 56127.99 N

$L\ cos[tex]L=5.61\times 10^4\ N$

(b) The net force acting in the x direction is :

$L\ sin\theta-R=0$

$R=L\ sin\theta$

$R=5.61\times 10^4\times \ sin(21)$

$R=2.01\times 10^4\ N$

Hence, this is the required solution.

Answer from: Quest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

Answer from: Quest

Newtons laws of motion

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