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Hoochie [10]
3 years ago
14

Choose the scenario in which the sound frequency of the waves is higher.

Physics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

B) the sound source moves towards you at 100 m/sec

Explanation:

The Dopper Effect is a phenomenon that occur when there is relative motion between an observer and a source of a wave. When this situation occurs, there is an apparent shift in frequency of the wave, as observed by the observer.

The apparent frequency observed by the observer is given by

f'=\frac{v\pm v_o}{v\pm v_s}f

where

f is the original frequency of the wave

f' is the apparent frequency

v is the speed of the wave

v_o is the velocity of the observer (positive if moving towards the source of the wave, negative otherwise)

v_s is the velocity of the source (negative if moving towards from the observer, positive otherwise)

In this problem, we want to find the scenario in which the sound frequency is higher.

We see that in all 4 scenarios, the sound source is moving: this means we have to find the scenario in which the denominator of the equation is smaller.

First of all, we notice the sound source moves towards the observer, v_s is negative, so the denominator is higher: this means that the correct option must be either A or B.

Also, we notice that since v_s is negative, a value larger in magnitude will mean a smaller denominator: therefore, the correct answer will be

B) the sound source moves towards you at 100 m/sec

Since this situation will make the denominator of the formula the smallest possible.

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Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\

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(b)

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3 years ago
Newton’s second law of motion states that when a force acts on an object, the object’s acceleration is equal to _____.
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3 years ago
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang
valina [46]

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

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Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta

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W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

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