The pressure at the bottom of a cylindrical container with a cross-sectional area of 45.5 cm2 and holding a fluid of density 420
1 answer:
Answer:
3.33 m
Explanation:
Pressure is the distributed force applied to the surface of an object per unit area. The force is applied perpendicular to the surface of the object. The SI unit of pressure is N/m² or Pa.
Hydrostatic pressure is the pressure that a fluid exerts at a point due to the force of gravity.
The relationship between pressure on the bottom of the container, atmospheric pressure and the pressure due to the depth of the fluid is given by:
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Answer:
Both of them reach the lake at the same time.
Explanation:
We have equation of motion s = ut + 0.5at²
Vertical motion of James : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
Vertical motion of John : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
So both times are same.
Both of them reach the lake at the same time.
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as
with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as
with ω denoting the angular velocity, which we are given. With that, the above becomes:
Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?
Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:
and so 45 rev/min = 4.71 rad/s.
A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.