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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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If you add 1 proton to Carbon, it will no longer be Carbon, it will be ________________.
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Nitrogen

Explanation:

Adding one proton to a carbon atom makes Nitrogen.

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The difference between an atom and another is the number of protons in them. This is the atomic number.

The periodic table of element is a list of elements arranged based on the number of protons they have. Every element on the table has unique number of protons which makes it differ from another.

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3 years ago
5. How much heat is generated when an
Mila [183]

Answer:

I HOPE THIS IS CORRECT

Explanation:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60oC

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60oC,Q=msΔT

S= specific heat =4200J/kgoC

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   2000×t=200×4200×60

      t=25200  

or   t=25.2×103sec.

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3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago
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