Answer:
C
technically B too but youre teachers not that smart so there you go
Answer:
Line the numbers from smallest to largest the subtract the smallest from the largest numbers.
Explanation:
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
