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3 years ago

As you jump on a pogo stick where is the potential energy the greatest?

Physics

1 answer:

slamgirl [31]3 years ago

7 0

The potential energy would be greatest at the highest point of the jump; the kinetic energy equaling 0.

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A plot of land has been surveyed for a new housing development with borders AB, BC, DC, and DA. The plot of land is a right trap

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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

$\sum F =0$

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

You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat

sweet [91]

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

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3 years ago

The amount of energy that remains after the amount of energy used for respiration is subtracted from the total is called

lawyer [7]

Answer:Gibb's free energy

Explanation:

The Free energy change describes the amount of energy that is available in any system to do work. It is often designated with the symbol G

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