It's a form of mechanical energy
Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg
Answer:
The magnitude of the acceleration is 
The direction is
north of east
Explanation:
From the question we are told that
The force exerted by the wind is 
The force exerted by water is 
The mass of the boat(+ crew) is
Now Force is mathematically represented as

Now the acceleration towards the north is mathematically represented as

substituting values


Now the acceleration towards the east is mathematically represented as

substituting values


The resultant acceleration is

substituting values


The direction with reference from the north is evaluated as
Apply SOHCAHTOA

![\theta = tan ^{-1} [\frac{a_e}{a_n } ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B%5Cfrac%7Ba_e%7D%7Ba_n%20%7D%20%5D)
substituting values
![\theta = tan ^{-1} [\frac{0.808}{1.269 } ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B%5Cfrac%7B0.808%7D%7B1.269%20%7D%20%5D)
![\theta = tan ^{-1} [0.636 ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B0.636%20%5D)
