<h2>
82.353 km/hr</h2>
Explanation:
The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.
The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.
Total displacement = ![135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East](https://tex.z-dn.net/?f=135%5Ctext%7B%20%7Dkm%5Ctext%7B%20%7DEast%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D0%5Ctext%7B%20%7Dkm%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D215%5Ctext%7B%20%7Dkm%5Ctext%7B%20%7DEast%5Ctext%7B%20%7D%3D%5Ctext%7B%20%7D350%5Ctext%7B%20%7Dkm%5Ctext%7B%20%7DEast)
Total time travelled = ![1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr](https://tex.z-dn.net/?f=1.5%5Ctext%7B%20%7Dhr%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D45%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D2.0%5Ctext%7B%20%7Dhr%5Ctext%7B%20%7D%3D%5Ctext%7B%20%7D90%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D45%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%2B%5Ctext%7B%20%7D120%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%3D%5Ctext%7B%20%7D255%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%3D%5Ctext%7B%20%7D4%5Ctext%7B%20%7Dhr%5Ctext%7B%20%7D15%5Ctext%7B%20%7Dmin%5Ctext%7B%20%7D%3D%5Ctext%7B%20%7D4.25%5Ctext%7B%20%7Dhr)
![\text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20velocity%20%3D%20%7D%5Cdfrac%7B%5Ctext%7BTotal%20displacement%7D%7D%7B%5Ctext%7BTotal%20time%20taken%7D%7D%3D%5Cdfrac%7B350%5Ctext%7B%20%7Dkm%7D%7B4.25%5Ctext%7B%20%7Dhr%7D%3D82.353%5Ctext%7B%20%7D%5Cfrac%7Bkm%7D%7Bhr%7D)
∴ Driver's average velocity = ![82.353\text{ }\frac{km}{hr}](https://tex.z-dn.net/?f=82.353%5Ctext%7B%20%7D%5Cfrac%7Bkm%7D%7Bhr%7D)
Answer:
2200000 = 2.2E6 min for light from Proxima to reach earth
8.3 min from light sun to reach earth
2.2E6/8.3 = 2.56E5 times for light from Proxima
Proxima is about 256,000 times farther away than the sun
Since the sun is about 93,000,000 = 9.3E7 miles from earth
Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away
Note - the speed of light is
3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law
![F=mg](https://tex.z-dn.net/?f=F%3Dmg)
![\dfrac{mv^2}{r}=mg](https://tex.z-dn.net/?f=%5Cdfrac%7Bmv%5E2%7D%7Br%7D%3Dmg)
![\dfrac{v^2}{r}=g](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%5E2%7D%7Br%7D%3Dg)
![v=\sqrt{rg}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Brg%7D)
Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula
![v=\sqrt{250\times9.8}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B250%5Ctimes9.8%7D)
![v=49.49\ m/s](https://tex.z-dn.net/?f=v%3D49.49%5C%20m%2Fs)
Hence, The speed of space station floor is 49.49 m/s.