7. If a load of 300N is pulled along the inclined plane shown in the figure, answer the following. B 200 N 0.5m 2m 300 N А i. Ca
1 answer:
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Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance, = 15.0 cm
image distance, = 5.0 cm
height of the object, = 9.0 cm
height of the image, = ?
Apply lens equation;
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
Answer:
Explanation:
Given that,
h(t) = -9.8t² / 2 + 125t + 500
for t > 0.
At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.
We want to find the maximum height reached by rocket
Using mathematics maxima and minima
let find the turning point when dh/dt = 0
dh/dt = -9.8t + 125
dh / dt = 0 = -9.8t + 125
9.8t = 125
t = 125 / 9.8
t = 12.76s
Let find the turning point to know if this time t = 12.76 is maximum or minimum point
Let find d²h / dt²
d²h / dt² = -9.8
Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.
Then, the maximum height reached is
h = -9.8t² / 2 + 125t + 500
h = -9.8(12.76)² / 2 + 125(12.76) + 500
h = -797.80 + 1595 + 500
h = 1297.2 m
The maximum height reached is 1297.2 m
From the attachment, the maximum height is 1297.2m at t = 12.76sec.
Comment, the result are the same for both the analysis aspect and the graphical aspect.
Answer:
757,93 feets
Explanation:
We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.
The angle of depression, its O in the graph.
Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.
ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).
Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.
Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.
We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:
This its equal to the distance from the boat to the shoreline.
