Answer:
low freezing point. high vapour pressure.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>
Answer:
Explanation:
Using the principle of moment, assuming the rod is uniform rod of mass 1 kg
the center of mass of the rod will be at 1 m
assuming the system is in equilibrium,
clockwise moment = anticlockwise moment
let the distance of the man shoulder be x from the center of gravity and also is the pivot point
total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and
2kg + 5 kg = 7 kg for front bucket
( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)
5 + 5 x + x = 7 - 7x
5 + 6x = 7 - 7x
6x + 7x = 7 - 5
13x = 2
x = 2 / 13 = 0.154 m
the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).
Actually what the problem meant about the westward
component of the ball’s displacement is the horizontal component of the
displacement. To help us better understand the problem, I attached a figure of
the situation.
We can see from the figure that to solve for the value of
the horizontal component, we have to make use of the sin function. That is:
sin θ = side opposite to the angle / hypotenuse of the
triangle
sin 42 = x / 40 m
x = (40 m) sin 42
x = 26.77 m
Therefore the ball has a westward
displacement of about 26.77 m
2.0y i think this is guess but if it right then thats good
