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
Answer:
All the competitors will move with the same velocity.
Explanation:
Here, the situations for each competitor are identical. Thus, they will exert the same force and hence, their velocities at each instants will be identical.
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law
Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula
Hence, The speed of space station floor is 49.49 m/s.
Answer: I am getting a result around the 54,800 area, so I am selecting the 55,000 HP answer.
