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Ad libitum [116K]

3 years ago

7

At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when th

e speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them?

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

(a): The frequency received by the observers is f'= 138,062.28 Hz.

(b): When the plane flies directly away from them, they receive a frequency of f'= 1772.46 Hz.

Explanation:

Vf= 333.33 m/s

Vo= 0 m/s

V= 342 m/s

f= 3500 Hz

(a) f' = f * ( V / (V - Vf) )

f'= 138062.28 Hz

(b) f'= f* ( V / (V - (- Vf) )

f'= 1772.46 Hz

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Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

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3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

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8. How did the measured angular magnification of the telescope compare with the theoretical prediction?

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

The focal length of B is $f_{objective } = 43.0 \ cm$

The focal length of A is $f_{eye} = 10.4 \ cm$

The theoretical angular magnification is mathematically represented as

$m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}$

$m = \frac{f_{objective }}{f_{eye}} = 4.175$

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range

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3 years ago

Which of the following processes are Exothermic? I. Melting II. Sublimation III. Vaporization IV. Condensation

mariarad [96]

Answer: condensation process is exothermic. The correct option is IV.

Explanation: Exothermic reaction is a chemical reaction in which heat is released to the surrounding environment.

condensation is defined as a process by which a medium changes from gaseous phase into liquid phase. Using water molecules to illustrate, to transform water molecules to gaseous phase it requires the heat of vaporization to be overcomed. This same hear of vaporization needs to be given off when changing back to liquid phase.

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Brilliant_brown [7]

Answer:

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Explanation:

As we know that the electrostatic force is a based upon inverse square law

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$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

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3 years ago