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Finger [1]
2 years ago
6

A fish is 1.2 m beneath the surface of a still pond of water. At what maximum angle can the fish look toward the surface (measur

ed with respect to the normal to the surface) in order to see a fisherman sitting on a distant bank? (for water, n = 1.333)
Physics
1 answer:
ivanzaharov [21]2 years ago
4 0

Answer:

49°

Explanation:

Depth at which the fish is found= 1.2 m

refractive index for water, n= 1.333

Using snell's law we can write

\frac{sin {i}}{sin {r}}=n

i and r are angle of incidence and angle of refraction respectively

Here, r= 90°

put value of n and r to get i

\frac{sin {i}}{sin {90}}=1.333

sin 90= 1  and sin^{-1}(1.333)= 49°

implies that i= 49°

hence, the maximum angle of incidence i= 49°

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Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors commit
Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

   = \dfrac{difference}{original}\times 100

   = \dfrac{63-42}{63}\times 100

   = \dfrac{21}{63}\times 100

   = 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0
2 years ago
A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h
babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

8 0
2 years ago
The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this
konstantin123 [22]

Answer:

3.81972\times 10^{-7}\ A

Explanation:

B = Magnetic field = 3\times 10^{-8}\ G

d = Diameter of loop = 16 cm

r = Radius = \frac{d}{2}=\frac{16}{2}=8\ cm

i = Current

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

The magnetic field of a loop is given by

B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A

The current needed to produce such a field at the center of the loop is 3.81972\times 10^{-7}\ A

5 0
2 years ago
A spring stores 10. joules of elastic potential
Mekhanik [1.2K]
The answer would be . Since we are looking for the spring constant you would need to use the formula
PEs =  \frac{1}{2} k {x}^{2}
. Then you'd substitute, for PEs and x.
10j=1/2k(.20m^2).
Then solve. k=500n/m
6 0
2 years ago
Read 2 more answers
A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0
2 years ago
Read 2 more answers
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