Answer:
C
Explanation:
I literally had this question yesterday
Brown bears can weigh half a ton (500 kilograms). The piston on the left has a cross-sectional
area of 1 square meter. The piston on the right has a cross-sectional area of 4 square meters.
How much force has to be exerted on the left piston to lift the bear? Show your work and
specify units.
In order to lift the Bear, a force of 1226.25 N has to be exerted on the left piston,
What is pressure is on the piston carrying the Bear?The pressure on the piston carrying the Bear is calculated as follows:
Pressure = force/areaForce of the Bear = 500 kg * 9.81 m/s² = 4905 N
Pressure on the right piston = 4905 N / 4 m² = 1226.25 N/m²
To lift the Bear, the pressure on both sides must be equal.
Force to be applied on the left piston = pressure * area
Force = 1226.25 N/m² * 1 m² = 1226.25 N
Therefore, a force of 1226.25 N has to be exerted on the left piston to lift the bear.
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Select the correct answer.
Which of these physical concepts is most important in civil engineering?
A energy transformations
B.motion under gravity
C. heat transfer
D. forces In structures
E. atomic structure
Answer:
option A energy transformation
A bridge a mass of 800 kg and is able to support up to 4 560 kg. What is its structural efficiency?
Answer:4560÷800=5.4
Explanation:
11. What are restrictions when building or completing a challenge?
Explanation:
The minimum exterior open spaces around buildings that are 55 metres or more, should be 16 metres. On sides where no habitable rooms face, a minimum space of 9 metres shall be left for heights above 27 metres.
Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.
a. M
b. L / T
c. M L / T^2
d. M / (L T^2)
e. No dimensions
Correct answer is option E. No dimensions
As we know formula Pressure (P) is [tex]\frac{F}{A}[/tex]
also,
Dimensional formula of Pressure is [tex]M^{1}L^{-1}T^{-2}[/tex]Dimensional formula of length is L Dimensional formula of mass is MDimensional formula of velocity is [tex]L^{1} T^{-1}[/tex]So, as given W=[tex]\frac{P*L^{3} }{M*V^{2} }[/tex]
Dimensional formula of W =[tex]\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }[/tex]
since all terms get cancelled
Work is dimensionless i.e no dimensions
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