Answer:
(θ) = 60°
Explanation:
Given:
Speed of canoe Vc = 2 m/s
Speed of River Vr = 1 m/s
Computation:
Vc (Cosθ) = Vr
2 (Cosθ) = 1
(Cosθ) = 1 / 2
(Cosθ) = (Cos60)
(θ) = 60°
Answer:
1.5 m
Explanation:
Length. L = 12 m
Width, W = 16 m
Area, A = 12 x 16 = 192 m^2
Let the width of pavement be d.
The new length, L' = 12 + 2d
the new width, W' = 16 + 2d
New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2
Difference in area = A' - A
285 = 192 + 56 d + 4d^2 - 192
93 = 56 d + 4d^2
4d^2 + 56 d - 93 = 0

\
d = 1.5 m
Thus, the width of the pavement is 1.5 m.
Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration
, so:

where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:

W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2
)= 1.976 rpm
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