Please help me asappppp!!

The diagram below shows two electromagnetic waves.

zaharov [31]

The best C) Wave 1 is generated by electric circuits and wave 2 is emitted by radioactive material.

The higher frequency waves will be radioactive material, while the electromagnetic radiation generated by circuits will have a much lower frequency.

3 0

4 years ago

What property is primarily responsible for determining the type of electromagnetic energy and peak wavelength emitted by a star

Galina-37 [17]

Answer:B. Temperature

Explanation:The temperature of the star such as Sun is measured. Using this measurement, its peak wavelength and energy can be determined.For determination of wavelength, Wien's displacement law is used. This law states that, the sun like body emits all kinds of wavelengths and thus is nearly a black body.For black body, the peak wavelength emitted is inversely proportional to the temperature of the body. From the wavelength, energy can be calculated.Temperature is the property which is primarily responsible for determining the type of electromagnetic energy and peak wavelength emitted by star.

5 0

3 years ago

A go-kart accelerates at a rate

stira [4]

Answer:

if it is at a speed of 12.5 m / s it would take 2 minutes

Explanation:

4 0

4 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$