Answer: the contents of this container weighs 4905 kg.m/s²
Explanation:
Given that;
volume of a container V = 0.5 m³
we know that standard gravitational acceleration g = 9.81 m/s²
specific volume of liquid filled in the container v = 0.001 m³/kg
now we express the equation for weight of the container.
W = mg
W = (pV)g
W = Vg / ν
so we substitute
W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg
W = 4.905 / 0.001
W = 4905 kg.m/s²
Therefore, the contents of this container weighs 4905 kg.m/s²
(b) Always act on the different bodies in opposite directions
Answer:
(2x + 4)(x - 4)=2x^2-4x-16
Longshore drift has a very powerful influence on the shape and composition of the coastline. It changes the slopes of beaches and creates long, narrow shoals of land called spits, that extend out from shore. Longshore drift may also create or destroy entire “barrier islands” along a shoreline.
Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So 
So 
v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
