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UkoKoshka [18]
3 years ago
11

The speed of light in water is 2.25 x 108 m/s. What is true about the index of refraction of water? A. It is less than 1. B. It

is more than 1. C. It is 0. D. It is equal to 1.
Physics
2 answers:
alexgriva [62]3 years ago
8 0
The answer is letter B. more then one. hope this helps!
guajiro [1.7K]3 years ago
4 0
Good afternoon!


We can see that \mathsf{2.25\times10^8\ \textless \ 3\times10^8 = c}

The index of refraction can be calculated as:

\mathsf{n = \dfrac{c}{v}}

As we noted, \mathsf{v \ \textless \  c} , and therefore:

\mathsf{\dfrac{v}{v} \ \textless \  \dfrac{c}{v}}\\ \\ \\ \mathsf{\dfrac{c}{v} \ \textgreater \  1}

Beccause n = c/v, we have:

n > 1

B - Is more than 1.
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A common flashlight bulb is rated at 0.23 a and 2.9 v (the values of the current and voltage under operating conditions). if the
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The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius
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3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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Answer:

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