The question on science please answer correctly

The frequency of a microwave is 1.2 x 10^9 hertz. what is the wavelength of the given problem.

Olenka [21]

Answer:

0.25 m

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field, oscillating in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

$c=3.0\cdot 10^8 m/s$

Microwave is an example of electromagnetic waves.

The relationship between wavelength and frequency for an electromagnetic wave is:

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

f is the frequency

For the microwave in this problem,

$f=1.2\cdot 10^9 Hz$

So its wavelength is

$\lambda=\frac{3.0\cdot 10^8}{1.2\cdot 10^9}=0.25 m$

7 0

3 years ago

A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of

kirza4 [7]

The answer would be 9.8 meters/ sec^2.

7 0

3 years ago

Read 2 more answers

HELP ASAP PLEASEEEEEEEEEEEE

ira [324]

It's the second graph!
it's the only one with a negative gradient.
so the temperature of the ball will fall in water as it looses its heat.

activate windows,:-P

8 0

3 years ago

A material made of _____ would be a very good conductor of energy.

aliya0001 [1]

Metal is a good conductor of energy

5 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago