A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Answer: 0.42 Amperes
Explanation:
Given that:
Current, I = ?
Electric charge Q = 100 coulomb
Time, T = 4.0 minutes
(The SI unit of time is seconds. so, convert 4.0 minutes to seconds)
If 1 minute = 60 seconds
4.0 minutes = 4.0 x 60 = 240 seconds
Since electric charge, Q = current x time
i.e Q = I x T
100 coulomb = I x 240 seconds
I = 100 coulomb / 240 seconds
I = 0.4167 Amperes (round to the nearest hundredth which is 0.42 amperes)
Thus, 0.42 Amperes of current flows in the circuit.
Answer:
one or more cells
Explanation:
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