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oee [108]
3 years ago
7

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

200(1 – 0.956e–0.18t ) where L(t) is the length (in centimeters) of a fish t years old.
(a) Find the rate of change of the length as a function of time
Physics
1 answer:
Ann [662]3 years ago
8 0

Answer:

Rate of change of length as a function of time is given by \frac{dL(t)}{dt}=34.416e^{-0.18t}

Explanation:

The length as function of time is given byL(t)=200(1-0.956e^{-0.18t})

Differentiating with respect to time we get

\frac{dL(t)}{dt}=\frac{d(200(1-0.956e^{-0.18t}))}{dt}\\\\\frac{dL(t)}{dt}=200\frac{d(1-0.956e^{-0.18t})}{dt}\\\\=200\times 0.18\times 0.956e^{-0.18t}\\\\\frac{dL(t)}{dt}=34.416e^{-0.18t}

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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed
Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

sin\theta = \frac{\lambda}{a}

              == \frac{633 *10^{-9}}{.360*10^{-3}}

           \theta = 0.100

y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m

with of central fringe  = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = w' = y_2 -y_1

calculation for y_2

sin\theta = 2\frac{\lambda}{a}

              = = 2*\frac{633 *10^{-9}}{.360*10^{-3}}

             \theta  = 2*0.100 = 0.200

y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m

w' = 0.0130  -6.54 *10^{-3}

w' = =6.46*10^{-3] m

6 0
3 years ago
Positive ions from a base and negative ions from an acid form a .
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Positive ions from a base and negative ion from an acid form salt.

 

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Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
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17.
avanturin [10]

Answer:

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