What type of modulation is typically used by broadcasting stations to transmit pictures on television screens?
2 answers:
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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.
Answer:
1.013 s
Explanation:
You can solve this problem using the equations for constant acceleration motion. The velocity at the bottom of the window can be found using this expression:
the gravity is negative as it opposes the movement.
Now, the time elapsed before the ball reappears is 2 times the time that it takes for the ball to go from the bottom of the window, reach maximum height, and reach again the bottom of the window, minus 2 times the time that it takes for the ball to travel from the top to the bottom of the window. The time that takes to the ball to reach maximum height, or in other words, to time that takes for the velocity of the ball to go from vo to 0m/s:
Then: