The mass of the Sikorsky UH-60 helicopter is 9300 kg. It takes off vertically at t = 0. The pilot advances the throttle so that the upward thrust of its engine (in kN) is given as a function of time in seconds by T=100+t2.
A.) How fast is the helicopter rising 3 s after it takes off?
B.) How high has it risen 3 s after it takes off?

Answers

Answer 1

The helicopter is rising at a rate of 0.0192 m/s after 3 seconds.

The helicopter has risen to a height of 43.9 m after 3 seconds.

A) To find the rate at which the helicopter is rising after 3 seconds, we need to calculate its acceleration using the formula:

a = F_net / m

where F_net is the net force acting on the helicopter, and m is its mass.

The net force is the difference between the upward thrust and the weight of the helicopter:

F_net = T - mg

where T is the upward thrust, g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and m is the mass of the helicopter.

Substituting the given values, we get:

F_net = (100 + [tex]3^2[/tex]) kN - 9300 kg x 9.81 m/[tex]s^2[/tex] = 178.7 kN

a = F_net / m = 178.7 kN / 9300 kg = 0.0192 m/[tex]s^2[/tex]

Therefore, the helicopter is rising at a rate of 0.0192 m/s after 3 seconds.

B) To find the height the helicopter has risen after 3 seconds, we need to integrate its velocity from t=0 to t=3, and add the initial height (which is zero):

v = ∫ a dt = ∫ (100 + [tex]t^2[/tex]) / m - g dt

v = (100/9300)t + (1/3)(1/9300)[tex]t^3[/tex] - 9.81t + C

At t=0, v=0, so C=0.

Substituting the given values, we get:

v = (100/9300)t + (1/3)(1/9300)[tex]t^3[/tex] - 9.81t

∫v dt = h = ∫ [(100/9300)t + (1/3)(1/9300)[tex]t^3[/tex] - 9.81t] dt

h = (50/9300)[tex]t^2[/tex] + (1/12)(1/9300)[tex]t^4[/tex] - 4.905[tex]t^2[/tex] + C

At t=0, h=0, so C=0.

Substituting t=3 seconds, we get:

h = (50/9300)([tex]3^2[/tex]) + (1/12)(1/9300)([tex]3^4[/tex]) - 4.905([tex]3^2[/tex]) = 43.9 m

Therefore, the helicopter has risen to a height of 43.9 m after 3 seconds.

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Related Questions

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Answers

Answer:

[tex]E=1.62\times 10^{14}\ J[/tex]

Explanation:

Given that,

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We need to find the energy obtained. The relation between mass and energy is given by :

[tex]E=mc^2[/tex]

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Answer:

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Answers

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A piece of gym equipment states that the maximum load it can hold is 300 kg. Why do you think it is important not to go over this limit?

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Answer:

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Explanation:

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A hockey puck, with an initial velocity of 65 km/h [W], ricochets off the boards. After 0.76 s in contact with the boards, its final velocity is 47 km/h [E]. Determine the acceleration of the puck.

Answers

Answer:

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Explanation:

For this exercise let's use the relationship between momentum and momentum variation

          I  = Δp

          F t = m v_f - mv₀

          F = m (v_f -v₀) / t

let's reduce the magnitudes to the SI system

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          v₀ = - 65 km / h = -18.056 m / s

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Answers

Answer:

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Important Tip:

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Explanation:

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http://www.sengpielaudio.com/calculator-period.htm

http://www.sengpielaudio.com/calculator-period.htm

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Answers

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114 minutes

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Answers

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Answers

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Answers

Answer:

it will option C ,hope it helps

Answer:

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Answers

Answer:

b

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Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10^-6 N. What electrostatic force will result if both charges are doubled and the distance remains the same?

Answers

Answer:

The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Explanation:

Coulomb's law of  electricity states that the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance of of separation between them.

In formula; F = KQ₁Q₂/d²

Using the formula for electrostatic force of attraction above to determine the force of between the two charged spheres

Let the charges be Q₁ and Q₂; distance of separation be d, K is a constant

Initially, F₁ = KQ₁Q₂/d² ---- (1)

F₁ = 3.0 * 10⁻⁶ N

when the charges are doubled, Q₁ = 2Q₁; Q₂ = 2Q₂; K and d remains constant

F₂ = 2KQ₁Q₂/d² ----(2)

Dividing equation (2) by (1) to find the ratio of their forces

F₂/F₁ = (2KQ₁Q₂/d²) / KQ₁Q₂/d²

F₂/F₁ = 2

Thus, F₂ is twice F₁.

Since F₁ = 3.0 * 10⁻⁶ N; F₂ = 2 * 3.0 * 10⁻⁶ N

F₂ = 6.0 * 10⁻⁶ N

Therefore, the electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Electrostatic force is a force imposed by one charge on another as a result of the field.Electrostatic force will  be 12.0 x 10⁻⁶ N if both charges are doubled and the distance remains the same.

What is the electrostatic force?

It is a force imposed by one charge on another as a result of the field.

The electrostatic force produced by one line charge on another line charge separated by distance d is determined by the charge potency of each charge as well as the separation distance between them.

Hence the electrostatic force is given by  

[tex]\rm F=\frac{Kq_1q_2}{d^2}[/tex]

The given data in the question ,

d is the distance between the charge

F₁ is the electric force for case 1= 3.0 x 10⁻⁶ N

F₂ is the  electric force for case 2= ?

Conditions for case 2;

(q₁=2q₁),(q₂=2q₂),d₂=d₁

For case 1,

[tex]\rm F_1=\frac{Kq_1q_2}{d_1^2}[/tex]

For case 2,

[tex]\rm F_2=\frac{K(2q_1)(2q_2)}{d_1^2}[/tex]

[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4F_1[/tex]

[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4\times3.0\times 10^{-6}\\\\\rm F_2=12.0\times10^{-6}[/tex]

Hence electrostatic force will be 12.0 x 10⁻⁶ N.  if both charges are doubled and the distance remains the same.

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https://brainly.com/question/9774180

The temperature in deep space is thought to be about 3K.what is 3K in degrees Celsius and in degrees Fahrenheit?

Answers

Answer:

c

Explanation:

bnbbbn nbnbnb bbbb bnbb

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