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
The work done by a constant force in a rectilinear motion is given by:

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

Therefore, the total work done is 578.123 J and the answer is option E
Answer: 75.02 m
Explanation:
u = 0 ( starts from rest )
v = 50 m/s
t = 3 s
( i ) a = v - u / t
= 50 - 0 /3
= 16.67
( ii ) s = ut + 1/2 at²
= 0 × 3 + 1/2 × 16.67 × 3 × 3
= <u>75.02 m</u>
Hope this helps...
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