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Ksivusya [100]
3 years ago
5

What is the mechanism of hydraulic machine ​

Physics
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

The hydraulic press allows the applied force (F1) to be converted into a higher force (F2) along some path as many times as the surface of the driven hydraulic cylinder (A2) is larger than the surface of the driving hydraulic cylinder (A1). The hydraulic press works on the basis of Pascal's law, which reads: External pressure is transmitted through the fluids equally in all directions.

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zhannawk [14.2K]

a) Cumulus is 100% the correct answer

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Answer:

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Explanation:

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3 years ago
A 70.0 kg person climbs stairs, gaining 3.90 meters in height. Find the work done (in J) to accomplish this task.
Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = 9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J

6 0
3 years ago
The lower the value of the coefficient of friction,the_the resistance to sliding​
malfutka [58]

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

               U  = \frac{F}{N}  

U is the coefficient of friction

F is the frictional force

N is the normal force

 We see that coefficient of friction is directly proportional to frictional force.

7 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
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