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Gwar [14]
2 years ago
13

As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?

Physics
1 answer:
Y_Kistochka [10]2 years ago
3 0

Answer:

C. Angle of Attack.

Explanation:

The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.

And thus, we have

Lift = Weight

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A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a
PSYCHO15rus [73]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

8 0
3 years ago
Imagine you are holding an apple. how could you increase the potential energy of this apple?
Ray Of Light [21]

If you're holding the apple at your waist, lift it to your mouth.
Potential energy relative to any level is proportional to its height
above that level.  Increase that height, and you've increased the
potential energy.

Since energy is conserved ... it never magically appears or
disappears ... you need to tell where that extra energy for the
apple came from.

It's exactly the work you did ... the force of your muscles acting
through the distance you raised the apple ... that became the
additional potential energy that the apple gained.

6 0
2 years ago
ANSWER ASAPPPPPPPPPPPP
Free_Kalibri [48]
My teacher said it was c

6 0
3 years ago
Read 2 more answers
) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th
Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

    m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w

Now, we will substitute 0 for both z_{1} and z_{2}, 0 for w, 334.9 kJ/kg for h_{1}, 2726.5 kJ/kg for h_{2}, 5 m/s for V_{1} and 220 m/s for V_{2}.

Putting the given values into the above formula as follows.

     m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w  

     1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0

                q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0
3 years ago
A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a
Scorpion4ik [409]

Answer:

176 min

Explanation:

456 g = .456 kg

Specific heat of ice s = 2093 J kg⁻¹

Heat required to raise the temperature by 25 degree

= mass x specific heat x rise in temperature.

= .456 x 2093 x 25

=23860 J

Heat required to melt the ice to make water at zero degree

= mass x latent heat

= .456 x 334 x 10³

=152304 J

Total heat required = 152304 + 23860 = 176164 J .

Time Required = Heat required / rate of supply of heat

= 176164 / 1000

176.16 min

4 0
2 years ago
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