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Westkost [7]
2 years ago
4

If a spear is jabbed exactly at that place where the fish appears in water, the fish is not killed. Give reason

Physics
2 answers:
murzikaleks [220]2 years ago
7 0

Answer:

The fish is not killed due to the Refraction (bending) of light.

Explanation:

The light experiences a phenomenon called Refraction that is the change of path or bending. As light travels from denser medium (water) to a less dense medium (air), its speed increases to some extent. Due to this change in speed of light, the path of the light changes. The observer sees the image of fish in the water formed by these bent rays of light coming off the fish which is at a different location and he/she jabs the spear at that location. However, as it is not the real location of fish but how it appears due to refraction, the spear is jabbed at a wrong location and the fish is not killed.    

Anestetic [448]2 years ago
3 0

Source:

ARC Centre of Excellence in Coral Reef Studies

Summary:

Fish are not as dumb as people sometimes think. Marine scientists have found that fish that are regularly hunted with spearguns are much more wary and keep their distance from fishers. In investigating the effects of marine areas closed to fishing by customary laws, an international team of researchers working in the Pacific found that fish exposed to speargun fishing take flight much earlier when a diver approaches compared with those living in protected zones.

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A uniform cylindrical steel wire (density: 7.8 x 103 kg/m3), 58.0 cm long and 1.34 mm in diameter, is fixed at both ends. To wha
lbvjy [14]

Answer:

T= 354.65 N

Explanation:

Given that

ρ= 7800 kg/m³

L= 58 cm

d=1.34 mm

f= 311 Hz

T= Tension

Speed of wave ,V

V = f λ      

V = f L

V= 311 x 0.58 m/s

V=180.38 m/s

Area of cross sectional

A= πr² mm²

A= 3.14 x 0.67² mm²

A=1.41 mm²

Mass = Density x Volume

m=7800\times 1.41\times 10^{-6}\ kg/m

m=0.0109 kg/m

Tension ,T

T=m V^2

T= 0.0109 x 180.38² N

T= 354.65 N

     

3 0
2 years ago
Select examples of simple harmonic motion that can be observed in everyday life.
liberstina [14]
A. a child gently swinging on a swing at small angles all the time
5 0
2 years ago
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False
Studentka2010 [4]
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
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4 0
2 years ago
Read 2 more answers
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6
Shalnov [3]

Answer:

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

Explanation:

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.

A) What is the wavefunction y(x,t) for the standing wave that is produced?

B) In which harmonic is the standing wave oscillating?

C) What is the frequency of the fundamental oscillation?

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. lambda=2L/n

when comparing the wave equation with the general wave equation , we get the wavelength to be

2*pi*x/lambda=6.98x

lambda=0.9m

we use the equation

lambda=2L/n

n=number of harmonics

L=length of string

0.9=2(1.35)/n

n=2.7/0.9

n=3

third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

8 0
2 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
2 years ago
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