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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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Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
7. A particle is forced to move in a straight line path. It returns to the starting point after 10 s. The total distance covered
lutik1710 [3]

d. All three statements are true.

8 0
1 year ago
A body weighing 108N moves with speed of 5m/s in a horizontal
poizon [28]

Answer:

i) 5 m/s^{2}  ii) 54 N  iii) 54 N

Explanation:

i) a = \frac{v^{2}}{r} ⇒ a = 5² ÷ 5 = 5 m/s^{2}

ii) m = \frac{W}{g} ⇒ m = 108 ÷ 10 = 10.8 kg , F = ma ⇒ F = 10.8 × 5 = 54 N

iii) F1 = F2 = 54 N

4 0
3 years ago
What is the KE if a 10kg mass traveling at 5 m/s
11111nata11111 [884]

Answer: KE = 25 J

Explanation: You must use the formula

KE = 1/2 m v²

to solve this problem.

KE = 1/2 (10 Kg) (5 m/s)

KE = 1/2 (50 kgm/s)

KE = 25 J

7 0
3 years ago
6th grade science I mark as brainliest.​
DerKrebs [107]

Answer:

2m 13\frac{1}{3}s

Explanation:

1.5m = 1s

200m = \frac{200}{1.5} × 1s

          = 133\frac{1}{3}s

          = 2m 13\frac{1}{3}s

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