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notka56 [123]
3 years ago
6

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.N·m(b) Find the angular acceleration of the airplane when it is inlevel flight.rad/s2(c) Find the linear acceleration of the airplane tangent to itsflight path.m/s2
Physics
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

_T}=24.57Nm

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

_T } = Fr

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

_T}=(0.795)*(30.9)

_T}=24.57Nm

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

_T}=I \omega

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

I = mr^2

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

_T}=I \omega

Making angular acceleration the subject of the formula; we have;

\omega = \frac{_T}{I}

ω = \frac{24.57}{707.51}

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a =  ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

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Answer:

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Explanation:

Acceleration = 5 m/s^2

Force = 10 N

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mass = force / acceleration

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mass = 2 kg

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Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

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4 0
3 years ago
a. What proportion of resistors have resistances less than 90 Ω? b. Find the mean resistance. c. Find the standard deviation of
Anettt [7]

Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

c) 9.43 Ω

d) 1

Explanation:

The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

4 0
3 years ago
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