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olya-2409 [2.1K]
3 years ago
6

How an elastic material can no longer be elastic but plastic instead?

Physics
1 answer:
Dvinal [7]3 years ago
3 0
<span><span>deformação elástica – é aquela em que removidos os esforços atuando sobre o corpo, ele volta a sua forma original</span><span>deformação plástica – é aquela em que removidos os esforços, não há recuperação da forma original.</span></span>
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Which of the following will cause a decrease in an object's weight? Increase in the mass of the object Decrease in the mass of t
Paul [167]
A decrease in mass will decrease an objects weight because 
weight = mass x gravitational constant
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A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain
irinina [24]

Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

If Object B is Negatively Charged, the Object A should be Positively Charged

If the Object B is Positively Charged, the Object A should be Negatively Charged

Sometimes it can Mix as a Neutral as well

Hope this Helps

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2 years ago
Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t
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A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b
netineya [11]

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

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h Sqrt(75)

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7 0
3 years ago
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
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