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wel
3 years ago
13

If the speed of light in air is 3.00 times 10 to the 8 m/s power and the speed of light in water is 2.25 times 10 to the 8 power

m/s what is the index of refraction of water

Physics
1 answer:
kykrilka [37]3 years ago
5 0
The refractive index is essentially the speed of light through a vacuum (extremely similar to air) divided by the speed of light through given medium.
I also attached some of my physics notes on refraction and that sort of jazz as they may help :)

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The speed of light in a vacuum is 2.998 x 108 m/s. what is its speed in km/h?
Artemon [7]
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7 0
3 years ago
Which insurance would contradict the big bang theory
n200080 [17]
The Big Bang theory suggests that the universe is constantly expanding and that stars, galaxies and other entities are moving away from each other. If there were a galaxy moving closer to earth then that would contradict the Big Bang theory since the entities should be moving away from earth and from each other.
7 0
3 years ago
(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force
balandron [24]

Answer:

The  force is  F =  1164.6\  lbf

The time is   \Delta t =  2.44 \  s

Explanation:

From the question we are told that

  The  mass of the car is  m  =  2500 \ lbm

   The  initial velocity of the car is u  =  25 \ mi/hr

   The final  velocity of the car is  v  =  50 \  mi/hr

  The acceleration is  a =  15 ft/s^2 =  \frac{15 *  3600^2}{ 5280} =  36818.2 \  mi/h^2

   

Generally the acceleration is mathematically represented as

      a =  \frac{v-u}{\Delta t}

=>   36818.2 =  \frac{50 - 25 }{ \Delta t}

=>   t = 0.000679 \  hr

converting to seconds

       \Delta t =  0.0000679 *  3600

=>     \Delta t =  2.44 \  s

Generally the force is mathematically represented as

        F  =  m * a

=>      F  =  2500 *  15

=>      F  =  37500 \ \frac{lbm *  ft}{s^2}

Now converting to foot-pound-second we have  

         F =  \frac{37500}{32.2}

=>        F =  1164.6\  lbf

7 0
3 years ago
Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a
worty [1.4K]

Answer:

Approximately 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}} assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity c and mass m. Increasing the temperature of this object by \Delta T would require Q = c\, m \, \Delta T.

Look up the specific heat of water: c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}.

It is given that the mass of the water in this mixture is m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}.

Temperature change of the water: \Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}.

Thus, the water in this mixture would have absorbed :

\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}.

Thus, the energy that water absorbed was: Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}.

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, Q(\text{water}), would be the opposite of the energy that the metal in this mixture released.

Thus: Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J} (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}.

Temperature change of the metal in this mixture: \Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}.

Rearrange the equation Q = c\, m \, \Delta T to obtain an expression for the specific heat capacity: c = Q / (m\, \Delta T). The (average) specific heat capacity of the metal pieces in this mixture would be:

\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}.

6 0
2 years ago
A ball is dropped from a height of 80m. The time, in seconds, it takes to reach the ground is:__________.
Aleonysh [2.5K]

Answer:

4.04 s

Explanation:

h = vi + 1/2 a t ^2

HERE  h = 80 m , vi = 0 , a =9.81 m/s^2

80 = 0 + 1/2 × 9.81 × t ^2

80 = 4.905 t^2

t^2 = 80/4.905

t ^2 = 16.30988

t = square root of 16.30988

t =  4.0385 s

t = 4.04 s

4 0
3 years ago
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