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2 years ago

Convert the speed of light 3.0x10^8 m/s to km/day

Physics

2 answers:

Aleksandr [31]2 years ago

7 0

Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

just olya [345]2 years ago

6 0

Answer:

$2592*10^7\frac{km}{day}$ .

Explanation:

We need to know that 1km = 1000 m= 10^3m and 1 day = 86400 = 864 * 10^2s.

$3*10^8 \frac{m}{s} = 3*10^8\frac{m}{s} \frac{86400 s}{1 day}$

$= 3*10^8*864*10^2\frac{m}{day} = 2592 * 10^{10} \frac{m}{day}$

Now, $2592 * 10^{10} \frac{m}{day}\frac{1km}{10^3m}$

$= 2592 * 10^{10}*10^{-3}\frac{km}{day} = 2592*10^7\frac{km}{day}$ .

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2 years ago

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2 years ago

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Sunny_sXe [5.5K]

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

n₁ sin θ₁ = n₂ sin θ₂

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In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

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kvasek [131]

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Maximum height reached = 35.15 meter.

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We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

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We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

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Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.