The point In the motion that the balls will be closest to each other is as at the time as the second ball is thrown or lunched.
<h3>Will the first ball always be travelling faster than the second ball?</h3>
No, The two motions are said to be parallel to one another, so they that they both the same timing. Hence, the time it takes for both balls to be able to fall to the ground is said to be the same.
Yes, the both balls have the same speed because of two balls are said to hit the ground, they are said to be done at the same height as well as the same energy and so they have the same speed.
Note that balls will take about 2 seconds to be able to hit the ground.
The horizontal projection velocity of the second ball can be changed so that the balls arrive at the ground at the same time and so the balls will be closest to one another if the second ball is thrown.
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This is not as simple as it looks.
His average speed is NOT (10km/hr + 50km/hr)/2 = 30 km/hr.
You have to use the definition of speed:
Speed = (total distance covered) / (time to cover the distance).
Let's say the distance up (and down) the hill is 'd' .
Then the time it takes to go up the hill is (d/10) hours.
And the time it takes to come down the hill is (d/50) hours.
Total distance = 2d km
Total time = (d/10) + (d/50) = (5d/50) + (d/50) = 6d/50
Speed = distance/time = 2d/(6d/50) = 100d/6d
<em>Speed = </em>100/6 = <em>16-2/3 km/hr</em>
400 * 3 = 1200
A jet can travel 1200 minutes for 3 seconds
<span>this is a dynamometer</span>
Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
