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kicyunya [14]
3 years ago
7

A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circ

ular path of radius R=9.33 mi and makes one revolution every T=0.123 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?
Physics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

The net force is 91780.8 N.

Explanation:

mass, m = 8950 kg

Radius, R = 9.33 miles = 15015.2 m

Time, T = 0.123 h = 442.8 s

There are two forces acting on the plane.

Horizontal force is the centripetal force and the vertical force is the weight.

Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N

The net force is

F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N

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Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

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Answer:

Explanation:

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