Answer:
jnfal4u4ryhfsbjls5
Explanation:
duehdakjweyedufkbshegygfr
Answer:
r=6.05km/hr
z=59.1 degree to the horizontal
Explanation:
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place
can be solved using pythagoras theorem
r^2=o^2+a^2
r^2=5.2^2+3.1^2
r^2=36.65
r=6.1km/hr is te birds velocity relative to the ground
tanz=5.2/3.1
z=tan^-1(5,2/3.1)
z=59.1 degree to the horizontal
Answer:
2m/s²
Explanation:
velocity = displacement (distance in a specified direction /time
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where
is the density of freshwater which has a constant value

Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as


So 
=> 
Now the work done is mathematically represented as

![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values

Force = work / dis
= 60/ 5
= 12 N