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jek_recluse [69]
3 years ago
12

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

at willbreak under a tension of 54N or more. He is eager to the fish to the dock in the shortest time possible. If the fish is at rest at the water's surface, what is the least amount of time in which the fisherman can raise the fish to the dock without losing it?
Physics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

a = 1 m/s^2

now for such acceleration we can use kinematics

d = \frac{1}{2}at^2

2 = \frac{1}{2}(1) t^2

t = 2 s

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Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

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where

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Here we have:

\lambda=632.8 nm = 632.8\cdot 10^{-9} m is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

n=520 lines/mm is the number of lines per mm, so the spacing between two lines is

d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m

Therefore, substituting m = 1, we find:

y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m

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Answer:

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