The net forward force =( 515 cos 31deg ) +( 926 cos 25 deg )
= 441.44N + 839.24 N
= 1280.7N in truck direction
If the velocity is decreased by half, then Kinetic energy will reduced to it's 4 times,
= 1.9 * 10³ / 4
= 4.75 * 10²
In short, Your Answer would be Option B
Hope this helps!
<span>. </span>What happens<span> to a</span>wave<span> as it moves into shallow water? When the water depth decreases to one half of a </span>wave's<span> wavelength, the </span>wave<span> starts to “feel the bottom”. ... On a gentle slope,</span>waves<span> begin to feel the bottom far from the </span>shore<span>.</span>
Answer:
The correct answer would be Saturn's Cassini Division.
Explanation:
Read about it here.
https://caps.gsfc.nasa.gov/simpson/kingswood/rings/
Hope this helps! :)
Oh what a sneaky problem !
The "range" of a projectile is how far out away from the gun it goes
before it falls to the ground.
The question is asking you: "What angle should you aim the gun
above the horizon so that the bullet doesn't go <em>any</em> distance away
from the gun, and it hits the ground exactly where the gun is "
And there are two different answers. Really sneaky !
Can you think of how you could fire the gun so that the bullet would
hit the ground right under the gun ? How about pointing the gun at
the ground ... aiming 90 degrees <em>below</em> the horizon ? That would
certainly do it. Is there another way to do it ?
How about aiming the gun straight up in the air ... 90 degrees <em>above</em>
the horizon ? Then it goes up, up, up, runs out of steam, stops rising,
starts falling, comes down, down, down, and hits the ground right where
the gun was.
As long as there's no air and no wind, the two angles are +90 degrees
and -90 degrees above the horizon.
' +90 above ' means straight up. ' -90 above ' means straight down.
