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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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olganol [36]

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2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

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The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

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2 years ago

Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol

ser-zykov [4K]

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PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

$t = \frac{d}{v}$