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rosijanka [135]
3 years ago
13

An incident light ray strikes water at an angle of 20 degrees. The index of refraction of air is 1.0003, and the index of refrac

tion of water is 1.33. The angle of refraction rounded to the nearest whole number is °.
Physics
1 answer:
WINSTONCH [101]3 years ago
5 0
In order to find the angle of the refracted ray, we may use the Snell's law (also known as Snell–Descartes law and the law of refraction). This law states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media such as water, glass, or air.
 In mathematical form, this is:

n₁sin(∅₁) = n₂sin(∅₂)
Where:n is the refractive index.

Plugging in the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91
The angle of refraction is 15°.
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A mover applies a net force of 28 N to a sofa that has a mass of 70 kg.
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Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

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The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I− (aq) The emf
Triss [41]

Answer: +2.10V

Explanation:

2Al(s)+3I_2(s)\rightarrow 2Al^{3+}(aq)+6I^-(aq)

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log K

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log [Al^{3+}]^2\times [I^-]^6

where,

E^o_{cell} = standard emf for the cell = +2.20 V

n = number of electrons in oxidation-reduction reaction = 6

E_{cell} = emf of the cell = ?

[Al^{3+}]  = concentration = 5.0\times 10^{-3}M

[I}^{-}]  = concentration = 0.10M

Now put all the given values in the above equation, we get:

E_{cell}=+2.20-\frac{0.059}{6}\log [5.0\times 10^{-3}]^2\times [0.10]^6

E_{cell}=2.10V

The standard emf for the cell using the overall cell reaction below is +2.10 V

5 0
3 years ago
A generator has a 100-turn coil that rotates in a 0.30-T magnitude B-field at a frequency of 80 Hz. The area of each loop is 0.0
loris [4]

Answer:

Explanation:

Flux Φ = nBA sinωt

t = 0 , flux = 0

dΦ / dt = nωBA Cosωt

e = nωBA Cosωt

= 100 x 2πx80 x.3 x .003 cos 2πx80 x .014

= 45.21 cos 7.0336

= 33.0668 v

= 33.1 V

4 0
3 years ago
A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3.4 kg weight is hanging on the string. The
Artyom0805 [142]

Answer:

5.45 J

Explanation:

When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system?

RKE = \frac{1}{2}I \omega ^2

where;

I = \frac{1}{2} mr^2

I = \frac{1}{2}*15*0.25^2

I = 0.46875 kg.m^2

\omega = \frac{1.0}{0.25}

\omega = 4 rad/s

RKE = \frac{1}{2}I \omega ^2

= \frac{1}{2} *0.46875*4^2

= 3.75 J

LKE = \frac{1}{2} mv^2

= \frac{1}{2} *3.4*1.0^2

= 1. 7 J

K.E = RKE + LKE

K. E = ( 3.75 + 1.7 ) J

K . E = 5.45 J

3 0
3 years ago
Read 2 more answers
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