Hey there!
The pressure under a liquid column can be , calculated using the following formula :
P = p x g x h
P atm = 1.013 x 10⁵ Pa
g = 9.8 m/s²
h = ?
h = P / ( p x g ) =
h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =
h = ( 1.013 x 10⁵ ) / ( 8820 ) =
h = 11.48 m ≈ 11.50 m
Hope this helps!
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:
C
Explanation:
Formula E=F/C also E=V/d
In this case use the second formula; E=V/d
Data given; E=4N/C d=8m
So v=E X d
V=4x8=32V
k.e=eV= 2X32=64eV
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components
Acceleration components
Time taken is 0.26088 seconds
The distance the beetle travels on the ground is 0.3677181864 m
Let
upthrust = T
weight = W = mg
Air resistance = F
When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)
Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)
Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
