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2 years ago

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Doppler Effect: A stationary source produces a sound wave at a frequency of 50 Hz. The wave travels at 100 feet per second. A ca

r is moving toward the sound source at a speed of 100 feet per second. What is the wavelength of the stationary sound source and the wave length that a person in the car perceives?
(Show the formulas and your calculations)

THANK YOU

1 answer:

Leno4ka [110]2 years ago

6 0

Answer:

︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎

︎

︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎

︎

︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎

Explanation:

Hope it helps!

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A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry

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Answer:

a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

$F_{e}$ - $T_{x}$ = m a

Axis y

$T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

Fe = $T_{x}$

T_{y} = W

Let us search with trigonometry the components of the tendency

cos θ = T_{y} / T

sin θ = $T_{x}$ / T

T_{y} = cos θ

$T_{x}$ = T sin θ

We replace

q E = T sin θ

mg = T cosθ

a) the electric force is

$F_{e}$ = q E

E = $F_{e}$ / q

E = -0.032 / 80 10⁻⁶

E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

q E / mg = tan θ

θ = tan⁻¹ qE / mg

Let's calculate

θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

θ = tan⁻¹ 0.3265

θ = 18 ⁰

sin 18 = x/0.30

x =0.30 sin 18

x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

$F_{e}$ = m aₓ

aₓ = q E / m

aₓ = 80 10⁻⁶ 4 10² / 0.01

aₓ = 3.2 m / s²

Axis y

W = m $a_{y}$

a_{y} = g

a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

a = √ aₓ² + a_{y}²

a = √ 3.2² + 9.8²

a = 10.31 m / s²

The Angle meet him with trigonometry

tan θ = a_{y} / aₓ

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-9.8) / 3.2

θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

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3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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2 years ago

A student compared some soccer players to the atoms in the gaseous state. Which of the following activities were the soccer play

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Answer:

The answer will be <em>D</em>

Explanation:

I choose answer <em>C</em> and got it wrong on the test. Also Liquid atoms vibrate fast and slide past each other. what answer <em>C </em>shows is a gas.

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3 years ago

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Suppose you are on an airplane moving at high speed. If you flip a coin straight up it will land in your lap rather than a great

maxonik [38]

It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.

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3 years ago

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Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

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3 years ago