Diagram A is the correct diagram
Answer:
25cm^2
Explanation:
area of square = side × side
length of side given = 5
area of this square = 5× 5
= 25cm^2
hope it helps
In order to determine the acceleration of the block, use the following formula:
Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:
Then, you have:
by solving for a, you obtain:
In this case, you have:
k: spring constant = 100N/m
m: mass of the block = 200g = 0.2kg
x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m
Replace the previous values of the parameters into the expression for a:
Hence, the acceleration of the block is 10 m/s^2
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:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s
