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pantera1 [17]
3 years ago
11

A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x

), with x in meters and F in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?
Physics
1 answer:
sattari [20]3 years ago
6 0
If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?
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If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?
Alisiya [41]

Answer:

The spring constant is 3750 N/m  

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

=>

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m  

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How many orbits/shells/energy levels does Rubidium have?
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The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
Nadusha1986 [10]

Answer:

The charge resides on the outer surface = 1.245 \times 10^{-12} C

Explanation:

Surface area of cell  (A) = 4.3\times 10^{-9}  m^{2}

Separation between two plate  (d) = 1.1 \times 10^{-8}  m  

Dielectric constant (k) = 4.2

Potential difference (\Delta V) = 85.7 \times 10^{-3} V

The capacitance of parallel plate capacitor in free space is given by,

           C = \frac{\epsilon_{o} A }{d}

Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

The Capacitance of capacitor is increase by k times when it placed in dielectric medium.

C_{dielectric}  = \frac{k \epsilon_{o} A }{d}

And we know that, C = \frac{Q}{ \Delta V}

So charge on the outer surface is given by,

      Q = \frac{k \epsilon A \Delta V }{d}

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      Q = 1.245 \times 10^{-12}

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3 years ago
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