All categories

3 years ago

A certain microwave has a

Physics

1 answer:

Tatiana [17]3 years ago

5 0

Answer:

$9.375\times 10^{9} Hz$

Explanation:

Frequency is mathematically defined as the quotient of speed divided by wavelength.

$Frequency= \frac {v}{W}$

where

v-is the speed of light

-w is wavelength.

Given the speed of the wave as $V=300000000 \ m/s$ and the wavelength $\lambda= 0.032\ m$ , we substitute these values in the Frequency function to solve for frequency:

$Frequency=\frac {300000000}{0.032}=9375000000 Hz=9.375\times 10^{9} Hz$

Hence, the wave's frequency is $9.375\times 10^9\ Hz$

You might be interested in

Margaret is the new director of research at a well-known pharmaceutical company. She has been asked to design a set of research

Serjik [45]

Answer:

Double blind experiment

Explanation:

It is an experimental method , which helps to avoid any impartiality and any error due to biasing .

The experiment give rise to very accurate results , which is very important for any experiment .

Hence , the new director Margaret , need to design a set of double - blind experiments.

7 0

4 years ago

Aang and Appa are flying over Ba Sing Se. They covered 500 m in 30 seconds while heading east. What was their velocity?

barxatty [35]

Answer:

Explanation:

I need help to marysol;)

8 0

4 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

How will heat flow between objects?

alukav5142 [94]

Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.

5 0

4 years ago

Read 2 more answers

When an object moves in a circular path, it accelerates toward the center of the circle as a result of ______.

kolbaska11 [484]

Answer:

Centripetal acceleration

Explanation:

An object moving around a xirxular path maintains its route as a result of centripetal force. However, its acceleration is caused by centripetal acceleration. Despite centripetal acceleration not being among the choices, it is the right answer.

Centripetal acceleration helps an object that navigates around a circular path to accelerate while centripetal force enables the movement of an object around a circular path to move inwards. Momentum, given as one of the choices is product of mass and velocity while friction is the force opposing movement of an object around a surface.

3 0

4 years ago