Please be so kind and patient to clarify this doubt:
I understand the part where mercury in column becomes heavier and goes down because of increased effective acceleration
What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being...
Please be so kind and patient to clarify this doubt:
I understand the part where mercury in column becomes heavier and goes down because of increased effective acceleration
What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being...
But why would we assume it otherwise ?
And my answer and explanation are sound if the air in the surrounding is not confined as in the elevator, right?
I do think so. ... because:
Mg / A = h(rho)g ---------------- static condition
M(g+a) / A = h'(rho)(g+a) ------------ accelerating condition.
ergo, h' = h
where rho is the density of mercury and M is the mass of the...
What would happen to the mercury in the column if a barometer is accelerated upwards?
In static condition, acceleration is g.
When the system moves upward with an acceleration a, effective acceleration = g + a.
The air above the mercury in the trough will press it with higher force...
What would happen to the mercury in the column if a barometer is accelerated upwards?
In static condition, acceleration is g.
When the system moves upward with an acceleration a, effective acceleration = g + a.
The air above the mercury in the trough will press it with higher force...
A bike starts from rest. It covers a distance of 1km from rest to rest. If the minimum time in which the bike covers the distance is 100s, the coefficient of friction between road and tyre = ?
s = (1/2)at^2
no idea how to go about it.