I am pretty sure that floodplains are most often found for rivers that exist on <span>
hilly areas at the base of mountains. In order to give yoy ans example which will make sure that this answer is quite a suitable one, nice example of f</span><span>loodplains</span>
is The Virgin River<span> at the upper end of Zion Canyon. It will definitely help you! Regards.</span><span>
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I'm sure you've noticed that an airplane high in the sky, far away
from you, looks like it's moving very slowly. At the same time,
somebody passing you on a skateboard whizzes past you at
high speed. The farther away something is from you, the slower
it appears to move.
The nearest star outside the solar system is almost 32 thousand times
as far away from us as the farthest visible planet (Saturn) is, and all of the
other stars are farther than that.
That's why you have to wait a few thousand years before you notice
that the shape of a constellation has changed.
To put it a slightly different way . . . Everything is in motion. The motion is
more noticeable for nearby things, and less noticeable for farther-away things.
Objects within our solar system are the only ones near enough so that a human
lifetime is a long enough period in which to notice the change in their position.
Even Pluto moves less then 1.5° against the 'background' stars in a whole year.
This all makes me feel small. How about you ?
Answer:
Explanation:
Given:
Force, f = 5 N
Velocity, v = 5 m/s
Power, p = energy/time
Energy = mass × acceleration × distance
Poer, p = force × velocity
= 5 × 5
= 25 W.
Note 1 watt = 0.00134 horsepower
But 25 watt,
0.00134 hp/1 watt × 25 watt
= 0.0335 hp.
Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
