All categories

3 years ago

Divergent faults form when tectonic plates _____.

Physics

1 answer:

Ronch [10]3 years ago

5 0

<span>move away from one another

(convergent is the opposite way)</span>

You might be interested in

I want ti know how to study

arlik [135]

Answer:

Make sure everything is organized have a planner it can help

Get rid of all distractions

Listen to music if it helps you concentrate

Have your notes

Being willing to stay focus on what you are doing

Understand what you are doing

And most off all Be Happy and Remain Calm : )

3 0

3 years ago

Read 2 more answers

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

Fofino [41]

Answer:

$5.04\cdot 10^8 A$

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

$\phi = 3.950\cdot 10^{-19}J$

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

$E=\frac{hc}{\lambda}$

where

h is the Planck's constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Using $E=\phi$ and solving for $\lambda$ , we find the maximum wavelength of the radiation that will eject electrons from the metal:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m$

And since

1 angstrom = $10^{-15}m$

The wavelength in angstroms is

$\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A$

3 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$