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erica [24]
3 years ago
10

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the pla

ne's velocity, v. The lift of a wing with an area of 190190 square feet is 95009500 pounds when the plane is going 220220 miles per hour. Find the lifting force on the wing if the plane slows down to 190190 miles per hour. (Leave the variation constant in fraction form or round to at least 5 decimal places.
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

F'=708.53 N

Explanation:

We have,

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane's velocity, v. It means tat,

F=kAv^2

k is constant

If, A = 190 Ft², v = 220 mph, F = 950 pounds

Let's find k first from above data. So,

k=\dfrac{F}{Av^2}\\\\k=\dfrac{950}{190\times 220^2}\\\\k=0.0001033

It is required to find the lifting force on the wing if the plane slows down to 190 miles per hour. Let F' is the new force. So,

F'=0.0001033\times 190\times (190)^2\\\\F'=708.53\ \text{pounds}

So, the lifting force is 708.53 pounds if the plane slows down to 190 miles per hour.

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A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
stepan [7]

Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

\rho = \frac{m_1}{V_1}

m_1 = \rho V_1 = 8.38 *2*10^{-3}

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m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

WE KNOW

PV = mRT

V, R and T are constant therefore we have

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\frac{p_2}{p_1}=\frac{m_2}{m_1}

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p_2 =7*10^{5} * \frac {.0159}{0.0167}

p_2 = 664081 N/m^{2}

8 0
2 years ago
I need help please, thank you!!!
stiks02 [169]

Answer:

135 mph

because 7+9=16

8 is left

if 7=40

9=50

then 8=45

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