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Gemiola [76]
3 years ago
7

An airplane is flying horizontally with a speed of 103 km/hr (278 m/s) when it drops a payload. The payload hits the ground 30 s

later. (Neglect air drag and the curvature of the Earth. Take g = 10 m/s².)
At what altitude H is the airplane flying?
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

H = 4500 m

Explanation:

  • Once dropped, the payload moves along a trajectory, that can be decomposed along two directions independent each other.
  • Just by convenience, we choose these directions to be coincident with the horizontal (-x) and vertical (y) axes.
  • As both movements are independent each other due to both are perpendicular, in the vertical direction, the initial speed is 0.
  • So, in order  to find the vertical displacement at any point in time, we can use the following kinematic equation, where a=-g., and H = -Δy.

        H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2}   = 4500 m

  • The airpane was flying at a 4500 m altitude.
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The rate of change of velocity is called:.
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Detailed Explanation:

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Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
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The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

  • Length of the incline, x_{max} = 1.536 m

Learn more about conservation of energy here:

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Explanation:

Given:

ω = 1 rev in 0.67 s

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ω = 2pi ÷ 0.67

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Rp = 10 cm

= 0.1 m

V = ω × r

= 9.38 × 0.1

= 0.938 m/s.

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