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Genrish500 [490]

3 years ago

13

james wants to lower the pitch coming from an electronic sound generator. how would he achieve that? a. he should increase the a

mplitude. b. he should increase the frequency c. he should decrease the amplitude. d. he should decrease the frequency. e. he should increase the intensity.

2 answers:

Gnesinka [82]3 years ago

3 0

Answer: Option D:<u> He should decrease the frequency. </u>

Pitch is a perceptual property of sound. It can be perceived in the sense of high or low musical melodies. The pitch and frequency are related. when frequency increases, pitch increases. And when frequency decrease, pitch decreases.

Hence, the correct option is D: he should decrease the frequency.

xxTIMURxx [149]3 years ago

3 0

He should decrease the frequency because frequency is the pitch of a sound so the lower the frequency the lower the pitch

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A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago

An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

$\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph$

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3 years ago

State the law of conversation of energy in your own words. Give an example of a situation that you have either encountered or kn

RSB [31]

That Energy Cannot Be Created Nor Destroyed

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3 years ago

Why does a Ray of light passing through a glass slab not shown dispersion?

Hatshy [7]

Answer:

After refraction at two parallel faces of a glass slab, a ray of light emerges in a direction parallel to the direction of incidence of white light on the slab. As rays of all colours emerge in the same direction (of incidence of white light), hence there is no dispersion, but only lateral displacement.

6 0

2 years ago

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl

ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
Both movements are independent each other, due to they are perpendicular.
In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

$v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

$t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

$v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

$\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.

4 0

3 years ago