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Aleks04 [339]
3 years ago
10

The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area

of 0.05m2.
Critical Thinking
Physics
2 answers:
kozerog [31]3 years ago
7 0

Answer:

50N

Explanation:

Pressure = Force/Area

Where P1 = P2

F1/A1 = F2/A2

Given that F1 =? A1 = 0.01m^2

F2= 250N A2= 0.05m^2

F1/0.01m^2 = 250N/0.05m^2

F1 × 0.05= 0.01 × 250

F1 = 0.01×250/0.05

F1 = 2.5/0.05

F1 = 50N

postnew [5]3 years ago
4 0

Answer:

Explanation:

Given that,

Small piston Hydraulic piston has an area

A1 = 0.01m²

If the force applied is 250N is applied to the small piston at an area of 0.05 m²

Then,

F2 = 250 N and A2 = 0.05m²

Then, applying pascal principle,

Pressure at small area = pressure are bigger area

P1 = P2

F1 / A1 = F2 / A2

F1 / 0.01 = 250 / 0.05

F1 / 0.01 = 5000

Cross multiply

F1 = 5000 × 0.01

F1 = 50 N

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Answer:

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Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

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Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

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Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

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Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

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Answer:

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m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

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