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Anna35 [415]
3 years ago
7

Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the

wing area is 16.6 m2, and the maximum lift coefficient is 2.1 with flaps down, calculate the stalling speed at sea level?
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

V=19.08 m/s

Explanation:

Airplane gross weight w=7780 N

Airplane wing area S=16.6 m²

Air density p=1.2250 kg/m³

Maximum lift coefficient CL=2.1

To find

Stalling Speed

Solution

The equation to find stalling speed is given below

W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\  V_{Stalling-speed}=19.08 m/s

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A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli
Vaselesa [24]

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

4 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
Which statement best describes metallic bonding
horrorfan [7]

Answer:

It's a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal bars. It can also be described as the sharing of free electrons among a structure of positively charged ions

4 0
3 years ago
A 0.14 -kg baseball is moving at 41 m/s. A 0.058-kg tennis ball is moving at 67 m/s. Which of the two balls has higher kinetic e
Iteru [2.4K]
Since kinetic energy is a form of energy using the equation KE=¹/₂mv², the units of measurement is in Joules (J). Therefore, the tennis ball had more kinetic energy than the baseball since velocity is a larger factor than the mass is when determining kinetic energy.

3 0
3 years ago
Read 2 more answers
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0
3 years ago
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