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djyliett [7]
2 years ago
7

A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of

the mass as it moves by. The speed is used to calculate the kinetic energy ( K E ) and the location of the photogates is used to calculate the elastic potential energy ( P E elastic ) . Interpret the chart to determine the missing K E and P E elastic values. Assume all values are exact.

Physics
1 answer:
svetoff [14.1K]2 years ago
4 0

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is  PE_{elastic} = 50J

The kinetic energy at point D is KE = 75J

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is E_ {m} = 100J

     So at point B

           E_{m} = PE_{elastic} +KE

           100 = PE_{elastic} + KE

   KE at point B is  50J

So     PE_{elastic} = 100 - 50 = 50J

     Now at point D

          E_{m} = PE_{elastic} +KE

    PE_{elastic} at point D is 25J

 So  KE = 100 - 25 = 75J

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Answer:

4.1\cdot 10^8 N

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

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F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N

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In a photoelectric effect experiment, electromagnetic radiation containing a finite distribution of wavelengths shines on a meta
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Answer with Explanation:

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A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with
strojnjashka [21]

Answer:

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Now, the horizontal and vertical components of these forces are:

F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1

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At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces F_{1y}\ and\ F_{2y} will be zero. This gives,

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Therefore,

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∴x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}

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2 years ago
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