C. length and mass
2 answers:
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Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.
Answer:
She will make the jump.
Explanation:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
First we will consider horizontal motion of stunt women
Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0
Substituting
So she will cover 77 m in 2.85 seconds
Now considering vertical motion, up direction as positive
Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 , time = 2.85
Substituting
So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.
So she will make the jump.