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Nikolay [14]
3 years ago
12

Why is the perfect mechine is not possible in the real life?​

Physics
1 answer:
Vinvika [58]3 years ago
3 0

Answer:

Because the mechanical advantage of the machine is affected by friction and weight but velocity ratio is not. So, mechanical advantage is less than velocity rate. Thus, the machine's efficiency is less than 100% and can't be a perfect machine

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A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
Georgia [21]

Answer:

Wp = 1,272 J

Wf = -940.8 J

Wn = 0

Wg = 0

Explanation:

  • Applying the definition of work, as the product of the component of the  force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
  • As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
  • We can get the work done by the force applied P, simply as follows:

        W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)

  • Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
  • The frictional force , can be written as follows:

        F_{fr} = \mu k * F_{n} (2)

        where μk = coefficient of kinetic friction = 0.24

         Fn = Normal Force

  • In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
  • Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
  • Replacing these values in (2), we get:

       F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N

  • Applying the definition of work, we can get the work done by the frictional force, as follows:

        W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ =  W_{ffr} = -940.8 N

4 0
3 years ago
After a 50-kg person steps from a boat onto the shore, the boat moves away with a speed of 0.70 m/s with respect to the shore. I
Svetradugi [14.3K]

Answer:

M=125 kg

v=1.75 m/s

Explanation:

From the law of linear momentum

  P =mv

Case 1     50*V =M* 0.7     equation 1

               50*V =(M+50)* 0.5    equation 2

equating 1 and 2

               M* 0.7=(m+50)* 0.5

               0.2 M= 25

                    M=125 kg

Putting value of M in equation 1

               50*V =125*0.7

                     V=1.75 m/s

                   

7 0
3 years ago
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shtirl [24]
5.6 g/ml. That is the density.
5 0
3 years ago
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Which statement is most likely correct?
DENIUS [597]

Answer:

the correct answer is B

8 0
4 years ago
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For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.
raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0
4 years ago
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