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Stels [109]
3 years ago
6

Watching a Helicopter Take Off At a distance of 58 ft from the pad, a man observes a helicopter taking off from a heliport. If t

he helicopter lifts off vertically and is rising at a speed of 45 ft/sec when it is at an altitude of 124 ft, how fast is the distance between the helicopter and the man changing at that instant? (Round your answer to one decimal place.)
Physics
1 answer:
kupik [55]3 years ago
5 0

To solve this problem we will apply the concepts related to classical mechanics, for which we will first consider the general equation that describes the movement of the body. From there, as we well know, we will obtain the derivative, which is equivalent to the speed of the body and the rate of change that is investigated in the problem.

Our values are given as,

h = 58ft

v = 45ft/s

h' = 124ft

We will start by defining the general height 'y' of the body at a reference distance of the lifting copter, given as

y^2 = (58)^2+(45t)^2

y^2 = 3364+2025t^2

Differentiating both the sides we have

2y \frac{dy}{dt} = 2025*2t

\frac{dy}{dt} = \frac{2025t}{y}

Now when the altitude of the copter is 124ft we have that

y^2 = (58)^2+(124)^2

y^2 = 18740

y = 136.89ft

And,

45t=124

t = 2.75s

Using these values at the derivative equation finally we have that

\frac{dy}{dt} = \frac{2025t}{y}

\frac{dy}{dt} = \frac{2025(2.75)}{136.89}

\frac{dy}{dt} = 40.68ft/s

Therefore the distance between the man and the helicopter is increasing at the rate of 40.68ft/s

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A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc
lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

  • <em>first charge, q₁ = 8.75 μC </em>
  • <em>second charge, q₂ = -6.5 μC  </em>
  • <em>electric field, E = 1.85 x 10⁸ N/C</em>
  • <em>distance between the two charges, r = 2.5 cm</em>

<em />

(a)

The attractive force between the charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N

The force on the negative charge due to the electric field is calculated as follows;

F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N

The tension on the wire is the resultant of the two forces and it is calculated as follows;

T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N

The force on the first negative charge due to the electric field is calculated as follows;

F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N

The force on the second negative charge due to the electric field is calculated as follows;

F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N

The tension on the wire is the resultant of the three forces and it is calculated as follows;

T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N

Learn more here:brainly.com/question/19565286

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A flywheel with a very low friction bearing takes 1.6 h to stopafter the motor power
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Answer:

(I). The initial rotation rate is 4.29 rad/s.

(II). The revolutions is 3932.

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Given that,

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(II). We need to calculate the revolutions

Using formula of revolutions

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\theta=\dfrac{24710}{2\pi}

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Hence, (I). The initial rotation rate is 4.293 rad/s.

(II). The revolutions is 3932.

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