Why are the surface features of the moon essentially unaltered since youth

1 answer:

8 0

Because there is no wind, rain, snow, sleet, or people to wear them down,
and no rivers, streams, or floods to cover them over.

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If the electron e − and the positron e + both have rest energy 0.51 MeV, nd the minimum energy a particle of light γ (a photon)

QveST [7]

Answer:

For Total energy and momentum to be conserved, the minimum energy of the photons released is equal to twice the rest mass energy of an electron that is $2 \times 0.51 MeV = 1.02 MeV$

The annihilation of electron -positron cannot produce a single photon. It is prohibited by the law of conservation of energy and momentum.

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Where are the three seismographs used to find the epicenter of this earthquake located?

iogann1982 [59]

The seismographs are located in Minneapolis, Detroit, and Charleston.

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The distance between two objects is increased by three times the oringinal distance. How will this change the force of attractio

tigry1 [53]

<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.

Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.

So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.

The forces become 1/16 of their original magnitude.<span>
</span>

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3 years ago

A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will

gtnhenbr [62]

1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
$x(t)=v_0t$
$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed $v_0=18.1 m/s$ , while the motion on the y-axis is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ , and initial height $h=1.85 m$ .

First of all, we can find the time t at which the ball reaches the ground by requiring $y(t)=0$ :
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
$x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m$

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
$v_x=18.1 m/s$
On the y-axis, the velocity is given by
$v_y(t)=gt$
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
$v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s$
And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

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A cyclist accelerates from a set of traffic lights.

Andrews [41]

Work equals Force times Distance. So Work = 250 Newton’s * 5 meters Work = 1,250

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