3 years ago

Example #2: A delivery truck drives 4 miles west before turning

1 answer:

4 0

The distance is 6+4 which =10 miles. The displacement would be 7.21 miles. This is because you have to use the Pythagorean theorem so a^2 + b^2 = c^2. So 6^2 + 4^2 = 52. Then take the square root and you get 7.21.

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How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase

morpeh [17]

Answer:

The value is $v_o = 5.488 \ m/s$

Explanation:

From the question we are told that

The emitted frequency increased by $\Delta f = 1.6 \% = 0.016 \$

Let assume that the initial value of the emitted frequency is

$f = 100\% = 1$

Hence new frequency will be $f_n = 1 + 0.016 = 1.016$

Generally from Doppler shift equation we have that

$f_1 = [\frac{ v \pm v_o}{v \pm + v_s } ] f$

Here v is the speed of sound with value $v = 343 \ m/s$

$v_s$ is the velocity of the sound source which is $v_s = 0 \ m/s$ because it started from rest

$v_o$ is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

$f_1 = [\frac{ v + v_o}{v + 0 } ] f$

=> $1.016 = [\frac{ 343 + v_o}{343 } ] * 1$

=> $v_o = 5.488 \ m/s$

8 0

3 years ago

The transfer of thermal energy via electromagnetic waves is know as what

katrin2010 [14]

Answer:

Radiation

Explanation:

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3 years ago

In a computer-based experiment to study diffraction, the width of the central diffraction peak is 15.20 mm. The wavelength of th

pogonyaev

Answer:

$a = 5.7 \times 10^{-5} m$

Explanation:

As we know that position of first minimum on the either side of central maximum is given as

$a sin\theta = \lambda$

$\theta =sin^{-1} \frac{\lambda}{a}$

so the width of the central maximum is given as

$W = L (2\theta)$

so we have

$15.20 \times 10^{-3} = 0.68 \times 2(sin^{-1} \frac{\lambda}{a})$

so we have

$0.011 = sin^{-1} \frac{\lambda}{a}$

$0.011 = \frac{638 nm}{a}$

$a = 5.7 \times 10^{-5} m$

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4 years ago

What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?

Delvig [45]

<em>im confused hold on imma send you a link to the answer</em>Explanation:

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3 years ago

What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Klio2033 [76]

Force =mass x acceleration =1100 x0.5 =550

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3 years ago