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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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In chemical reactions, energy is absorbed or released in the form of:
disa [49]

Answer:

Heat

Explanation:

Because chemical energy is stored, it is a form of potential energy. When a chemical reaction takes place, the stored chemical energy is released. Heat is often produced as a by-product of a chemical reaction – this is called an exothermic reaction.

Hope this helped.

3 0
3 years ago
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A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long
steposvetlana [31]
Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2 

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared 

vf^2=v0^2 + (2)(g)(yf-y0) 
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m) 
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2) 
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8 0
3 years ago
8.A catcher puts out his mitt and catches the BASEBALL BALL. Is the work
nika2105 [10]

Answer:

The catcher does negative work on the ball because the force exerted by the catcher is opposite in direction to the motion of the ball.

Explanation:

7 0
3 years ago
Please hurry this is urgent right answers only!!!!!
dem82 [27]

Explanation:

if two individual forces are of equal magnitude and opposite direction, then the force is said to be balanced. when there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction.

6 0
3 years ago
A projectile is launched from ground level with an initial speed of 53.7 m/s. Find the launch angle (the angle the initial veloc
Simora [160]

Answer:\theta =75.52^{\circ}

Explanation:

Given

initial speed of Launch(u)=53.7 m/s

Range of Projectile =Maximum height of Projectile

Range is given by R=\frac{u^2\sin 2\theta }{g}

Maximum height is given by H_{max}=\frac{u^2\sin ^2\theta }{2g}

R=H_{max}

\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin ^2\theta }{2g}

\frac{u^2\sin 2\theta }{g}-\frac{u^2\sin ^2\theta }{2g}=0

\frac{u^2\sin \theta }{g}\cdot \left [ 2\cos \theta -\frac{1}{2}\right ]=0

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2\cos \theta -\frac{1}{2}=0

\cos \theta =\frac{1}{4}

\theta =75.52^{\circ}

6 0
3 years ago
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