How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
Well,
When an object's velocity changes, we call it acceleration.
Acceleration: The time rate of change in an object's velocity
Total momentum after the collision: +200 kg m/s to the right
Explanation:
We can answer this question by using the law of conservation of momentum, which states that for an isolated system (=no external forces acting on the system), the total momentum is conserved.
Mathematically,
where
is the total momentum before the collision
is the total momentum after the collision
In this problem, the system consists of two hockey players. Before the collision, their total momentum is
(to the right)
Therefore, according to the law of conservation of momentum, their total momentum after the collision must be the same:
And given that the sign is +, the direction is still the same, therefore to the right.
Learn more about momentum:
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brainly.com/question/6573742
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Answer:
(a) 0.942 m
(b) 18.84 m/s
(c) 2366.3 m/s²
(d) 0.05 s
Explanation:
(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.
(b) Its frequency, f, is 1200 rev/min = 
rev/s = 20 rev/s.
Its angular frequency, ω = 2πf = 2π × 20 = 40π
The speed is given by
v = ωr = 40π × 0.15 = 6π = 18.84 m/s
(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²
(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.
T = 1/20 = 0.05 s
Answer:
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
Explanation:
Newton’s third law motion states that for every action there will be an equal and opposite reaction.
Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.
Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.
Uses of thrust reversal in practice:
When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
