Answer:
A.The vertical velocity is constantly increasing as the ball falls.
B.The horizontal velocity does not noticeably change as the ball falls.
G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.
H.The velocity vector of the ball changes as it travels through the air.
Explanation:
As the ball is projected horizontally so here the vertical component of the velocity is zero
So the time to reach the ground is given as

so we will have

so this is the same time as the ball is dropped from H height
Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.
So overall the velocity vector will change due to net acceleration g
From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560
let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs
(2400-640t) / 640 = 640t / 560
560(2400 - 640t) = 640t x 640
t = 1.75hrs
Constant velocity means the netto force = 0, therefore F(gravity) = F(astronaut).
175N divided by 87,5kg = 2.00kg/N
Answer:
Properties of semiconductors are determined by the energy gap between valence and conduction bands. To understand, what is semiconductor, we have to define these terms. In solid-state physics, the energy gap or the band gap is an energy range between valence band and conduction band where electron states are forbidden.
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<span><span>The
best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V
Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V
Since Cu is acting as the anode, the equation needs to be
reversed.
Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V
Ecell= -2.37 V+ (- 0.34 V) = -2.71 V
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</span><span>Hope my answer would be a great help for you. </span> </span>
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