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3 years ago

Explain what happens to the pitch of a cell phone ring when the wavelength of a sound wave increases

Physics

2 answers:

max2010maxim [7]3 years ago

4 0

You get a more low sound.

Conversely, when the wavelength becomes shorter you get a more treble sound.

;-)

andrew-mc [135]3 years ago

4 0

"Pitch" of sound is what we hear as the result of its frequency.

When the wavelength of any sound INcreases, its frequency DEcreases.

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Explain why objects moving in fluids must have special shapes ?

Alexus [3.1K]

Explanation:

Fluids exert both drag and lift forces on moving objects. Drag is the frictional force opposing motion. Lift is the force perpendicular to motion.

Some objects, like parachutes, are designed with large cross sectional areas to increase drag force. Usually though, objects are designed to minimize drag force. It's why cars, planes, and boats have sleek shapes.

Airplane wings have shapes called airfoils that generate lift. It's what makes them fly. The same shape is found in racecar spoilers. These spoilers use lift force to push down on the rear tires, increasing traction.

8 0

3 years ago

What is the net displacement of the particle between 0 seconds and 80 seconds?

Illusion [34]

After a thorough research, there exists the same question that has choices and the link of the graph (http://i37.servimg.com/u/f37/16/73/53/52/graph410.png)

<span>Choices:
A. 160 meters
B. 80 meters
C. 40 meters
D. 20 meters
E. 0 meters
</span>
The correct answer is letter E. 0 meters.

7 0

3 years ago

Which type of rock can only form below earths surface?

Inessa [10]

Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous

Explanation:

7 0

1 year ago

Read 2 more answers

One of the advantages of alternating current (ac) over direct current (dc) is the ease with which voltage levels can be increase

Svet_ta [14]

Answer:

True

Explanation:

In an alternating current, voltage levels can be easily increased or decreased as per the requirements of the energy distribution in practical world.

Hence, the given statement is true

3 0

2 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

3 years ago