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Ber [7]
3 years ago
13

A 4600 kg helicopter accelerates upward at 2.0 m/s squared by the upward life (Fa) exerted by the air on the propellers. Assume

that there is no air resistance.
Calculate the net force of helicopter using Newton's 2nd Law.
Calculate the helicopter's weight?
Calculate the lift-force (Fa) exerted by the air on the propellers
Please I need the answers
Physics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

(a) The net force upward force of the helicopter is -35,880 N.

(b) The weight of the helicopter is 45,080 N.

(c) The lift-force exerted by the air on the propellers is 9,200 N.

The given parameters:

  • <em>mass of the helicopter, m = 4600 kg</em>
  • <em>acceleration of the helicopter, a = 2 m/s²</em>

The net force upward force of the helicopter is calculated as follows;

F = ma - mg\\\\F = m(a - g)\\\\F = 4600(2 - 9.8)\\\\F = - 35,880 \ N

The weight of the helicopter is calculated as follows;

W = mg\\\\W = 4600 \times 9.8\\\\W = 45,080 \ N

The lift-force exerted by the air on the propellers is calculated as follows;

F = ma\\\\F = 4600 \times 2\\\\F = 9,200 \ N

Learn more about Newton's 2nd Law here: brainly.com/question/3999427

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A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of
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The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

  • The peak emf produced in a rotating loops= N×B×A×w
  • N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
  • So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
  • N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
  • Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
  • Now, B= 24/(300×0.00261×42)

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

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6 0
2 years ago
6) A penny is dropped from the roof of a house. The penny is traveling at 35 m/s right before it hits
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8 0
3 years ago
A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.
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Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

8 0
3 years ago
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