Answer:
The actual elevation angle is 12.87 degrees
Explanation:
In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that .
Using Snell's Law we can write:
,
Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.
We thus have:
. Calling
the actual angle of elevation, we get from the picture that
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula
A negative result has no sense, therefore the rowing speed in still water was 3 km/h