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3 years ago

What is the wavelength of a 2.50-kilohertz sound

1 answer:

3 0

The formula v=fλ can be used here.

326=2500*λ

Note the 2500 as 2.5kHz is 2.5 thousand Hz.

λ = 326/2500
= 0.1304m = 0.130m

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A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is

zvonat [6]

Work done is when a force is exerted to cause a displacement in a certain object.
the equation for work done ;
work done = force applied * displacement of the object
when the force applied is not in the same direction as that of the displacement of the object then the effect of the force is not its whole value. The force is then applied at an angle to that of the displacement of the object, then the resultant force is the force exerted* cos of the angle between force and displacement, in this instance the angle is 40 °.
the new equation is then;
work done = force cos 40° * displacement
after substitution,
work = 25 N * 0.76 * 50 m
= 957.55 J
round it off
= 9.6 *10² J
the correct answer is B

5 0

3 years ago

Read 2 more answers

I drove from Oklahoma to California (1452.9 miles) in 23 hours? What was my average speed?

storchak [24]

Answer:

63 miles per hour

Explanation:

you just divide miles and hours to get your average speed

5 0

2 years ago

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?

AnnZ [28]

V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
F = 0.08810409956 Hz

4 0

3 years ago