Answer:
The magnitude of the tension in he string is equal to the magnitude of the weight of the object.
Explanation:
According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.
In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,
The tension in string(T) , The weight of the object(W) .
∴ F = 0
T - W = 0
So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of the object.
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
Answer:
momentum=mass x velocity= 10 x 2 = 20kgm/s
