Answer:
a) 10.29° upstream
b) t=338.7s
Explanation:
If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:
The realitve velocity of the boat respect to the water is:
![V_{B/W}=[3*cos\beta ,3*sin\beta ]](https://tex.z-dn.net/?f=V_%7BB%2FW%7D%3D%5B3%2Acos%5Cbeta%20%2C3%2Asin%5Cbeta%20%5D)
where β is the angle it has to be pointed at.
From the relative mvement equations:
where ![V_B=[V*cos\alpha ,-V*sin\alpha ]](https://tex.z-dn.net/?f=V_B%3D%5BV%2Acos%5Calpha%20%2C-V%2Asin%5Calpha%20%5D)
From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:
Solving for V:
V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:
The amount of time:
I really need these points thx a lot
Answer:
The answer is 1.87nm/s.
Explanation:
The 
water loss must be replaced by of sap. 110g of sap corresponds to a volume of
thus rate of sap replacement is
The volume of sap in the vessel of length 
is
,
where 
is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
taking the derivative of both sides we get:
on the left-hand-side 
is the velocity 
of the sap, and on right-hand-side 
; therefore,
and since the cross-sectional area is
;
therefore,
solving for we get:
which is the upward speed of the sap in each vessel.
Answer:
0.83m/s
Explanation:
Given parameters:
distance = 1000m
time taken = 20min
Unknown:
Speed of the walking person = ?
Solution:
To solve this problem;
Speed = 
Time taken should be in seconds;
1 min = 60s
20min = 20 x 60 = 1200s
Speed = 
= 0.83m/s
