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kkurt [141]
3 years ago
7

How do streams flow?

Physics
2 answers:
salantis [7]3 years ago
5 0

Answer: When rain falls on the land, it either seeps into the ground or becomes runoff, which flows downhill into rivers and lakes, on its journey towards the seas. As small creeks flow downhill they merge to form larger streams and rivers. Rivers eventually end up flowing into the oceans.

Explanation:

AfilCa [17]3 years ago
3 0
When rain falls on the land, it either seeps into the ground or becomes runoff, which flows downhill into rivers and lakes.
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When a hot and cold object are placed in contact, the hot one loses energy. Does this violate energy conservation? Why or why no
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Answer:

This does not violate the conservation of energy.

Explanation:

This does not violate the conservation of energy because the hot body gives energy in the form of heat to the colder body, this second absorbs energy. This will be the case until both bodies reach the same temperature, reaching thermal equilibrium and reducing the transfer of thermal energy. In this way the energy was only transferred from one body to another but the total energy of the system (body 1 plus body 2) will be the same as in the beginning, respecting the principle of conservation of energy or also called the first principle of thermodynamics .

The part of physics that studies these processes is in turn called heat transfer or heat transfer or thermal transfer. Heat transfer occurs whenever there is a thermal gradient or when two systems with different temperatures come into contact. The process persists until thermal equilibrium is reached, that is, until temperatures are equalized. When there is a temperature difference between two objects or regions close enough, the heat transfer cannot be stopped, it can only be slowed down.

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Is a mirrors surface opaque or transparent or translucent?
dangina [55]
A mirror is opaque you can not see through it but you can see a reflection within it
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When an object moves away from Earth, its light waves are stretched to a lower frequency, or longer wavelength, and thus we say
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Your answer will be False because wavelengths are usually are use as sound waves. 
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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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