In a hydraulic system, a force of 400 N is exerted on a piston with an area of 0.001 m2. The load-bearing piston in the system has an area of 0.2 m^2. Required:
a. Calculate in kPa the pressure in the hydraulic fluid induced by the applied pressure.
b. What is the magnitude of the force exerted on the load bearing piston by the hydraulic fluid?

Answers

Answer 1
Answer:

Answer:

Explanation:

Pressure on the hydraulic system is expressed as;

Pressure = Force/Area

Given

Force on the fluid = 400N

Area = 0.001m²

Pressure in the fluid = 400/0.001

Pressure in the fluid = 400,000N/m²

1N/m² = 0.001kPa

400,000N/m² = x

x = 400,000 × 0.001

x = 400kPa

Hence the pressure in kPa is 400kPa

b) Using the formula;

Pa = Pb

Fa/Aa = Fb/Ab

Pa = Fb/Ab

Fb = PaAb

Fb = 400,000(0.2)

Fb = 80,000N

Hence the magnitude of the force exerted on the load bearing piston by the hydraulic fluid is 80,000N

Answer 2
Answer:

Final answer:

In a hydraulic system, a force exerted creates a pressure that is transmitted equally throughout. The induced pressure in the hydraulic system is 400 kPa. By applying the same pressure across the larger piston (0.2 m2), a force of 80,000 N is generated on the load-bearing piston.

Explanation:

The subject of this question is the physics topic of hydraulic systems, specifically how forces and pressure interactions. Use Pascal's principle which states that a change in pressure at any point in an enclosed fluid under equilibrium will be transmitted equally to all parts of the fluid.

a. To find the pressure in kPa induced by the applied pressure, use the formula P=F/A, where P is pressure, F is force and A is area. With the force F = 400 N and area A = 0.001 m2, the induced pressure is P = 400 N / 0.001 m2 = 400,000 Pa or 400 kPa.

b. The force exerted on the load-bearing piston by the hydraulic fluid is calculated by rearranging the formula to F=PxA. So, F = 400kPa x 0.2m2 = 80,000 N. Therefore, the force exerted on the load-bearing piston by the hydraulic fluid is 80,000N.

Learn more about Hydraulic Systems here:

brainly.com/question/33286738

#SPJ3


Related Questions

Lab: Weather Patterns
An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?
Which if, any, of these statements are true? (More than one may be true.) Assume the batteries are ideal. Check all that apply. A battery supplies the energy to a circuit. A battery is a source of potential difference; the potential difference between the terminals of the battery is always the same. A battery is a source of current; the current leaving the battery is always the same.
What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m  k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 516 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.6 ft deep, estimate the magnitude of his average acceleration as he slowed to a stop (that is while he was impacting the snow).

Answers

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 32.174* 516+0^2)\n\Rightarrow v=182.218\ ft/s

v^2-u^2=2as\n\Rightarrow a=(v^2-u^2)/(2s)\n\Rightarrow a=(0^2-182.218^2)/(2* 3.6)\n\Rightarrow a=-4611.58\ ft/s^2

Magnitude of acceleration while stopping is 4611.58 ft/s²

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer:

The correct option is  D

Explanation:

From the question we are told that

  The maximum electric field strength is  E = 1.0 *10^(11) \  V/m

   The  area is  A = 5.0 \ mm^2  = 5.0 *10^(-6) \  m^2

Generally the force the laser applies is mathematically represented as

       F = \epsilon_o * E ^2 * A

Here  \epsilon_o = 8.85*10^(-12) C/(V \cdot m)

      F =  8.85*10^(-12)  * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )

=>   F =  4.43 *10^(5) \ N

1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

F_(net)=m\,a

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg

Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.

Answers

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

W = U_f - U_i

Where,

W = Work

U_f = Final gravitational Potential energy

U_i = Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

U_i = (-GMm)/(R)

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is R_(orbit) \approx R then the energy possessed by the spacecraft is

U_f = (-GMm)/(2R)

Work needed to put it in orbit is the difference between the above two

W = U_f - U_i

W = -GMm ((1)/(2R)-(1)/(R))

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

W = (GMm)/(2R)

A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer:

The value is p =   - 2.63 \ Diopters

Explanation:

From the question we are told that  

      The value of the far point is  a =  40 \ cm  =  0.4 \  m

      The distance of the lens to the eye is  b =  2 \ cm = 0.02

Generally

        1 Diopter = >  1 m^(-1)

Generally the power spectacle lens needed is mathematically represented as

           p = (1)/(d_o )  + (1)/(d_i)

Here d_o is the object distance which for a near sighted person is d_o =  \infty

And  d_i is the image distance which is evaluated as

        d_i =  b - a

=>     d_i =  0.02 - 0.4

=>     d_i = -0.38 \  m

So

         p = (1)/(\infty )  + (1)/(-0.38)

=>      p = 0   - 2.63

=>      p =   - 2.63 \ Diopters