Investigating energy waves other than light is usually accomplished by using a ____
1 answer:
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The acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
The given parameters;
- <em>initial velocity of the engine, u = 1341 m/s</em>
- <em>final velocity of the engine, v = 7600 m/s</em>
- <em>time of motion, t = 2 minutes = 2 x 60 s = 120 s</em>
The acceleration of the SRB and main engine is calculated as follows;
Thus, the acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
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THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4 (1)
Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4 (2)
Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4 (3)
To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8 (4)
Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0
Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).
From (3), obtain
a = -4/4.142² = -0.2332
The equation for the green hose is
y = 0.2332(x - 4.142)² + 4
THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.
A graph of y(x) for both hoses is shown in the attached figure.
Answers:
a. The red hose will throw the water higher.
b. The equation for the green hose is
y = -0.2332(x - 4.124)² + 4,
with the origin at a height of 4 feet above ground level.
c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
The corresponding range is -4 ≤ y ≤ 4 feet.
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4 (1)
Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4 (2)
Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4 (3)
To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8 (4)
Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0
Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).
From (3), obtain
a = -4/4.142² = -0.2332
The equation for the green hose is
y = 0.2332(x - 4.142)² + 4
THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.
A graph of y(x) for both hoses is shown in the attached figure.
Answers:
a. The red hose will throw the water higher.
b. The equation for the green hose is
y = -0.2332(x - 4.124)² + 4,
with the origin at a height of 4 feet above ground level.
c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
The corresponding range is -4 ≤ y ≤ 4 feet.