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Brilliant_brown [7]
3 years ago
11

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

d Jerry are using the same force F. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do.
Physics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

W = F\cdot \Delta s

Where:

W - Work, measured in joules.

F - Force exerted on the object, measured in newtons.

\Delta s - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

W_{net} = W_{Joel} - W_{Jerry}

W_{net} = F\cdot \Delta s - F\cdot \Delta s

W_{net} = (F-F)\cdot \Delta s

W_{net} = 0\cdot \Delta s

W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

W_{net} = W_{Joel}

W_{net} = F\cdot \Delta s

Hence, the work done by Joel is greater than the work done by Jerry.

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Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

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The angle that locates the first dark fringe is given by :

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sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

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sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

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What are the units, if any, of the particle in a box wavefunction. What does this mean?
SIZIF [17.4K]

Answer:

  • [\psi]= [Length^{-3/2}]
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Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

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As the differentials has units of length

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for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

[\psi]^2 = [Length^{-3}]

taking the square root this gives us :

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learn more about acetic acid, visit;

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Since those constants are proportional to the magnitude of the forces:

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