Answer:
A lens placed in a transparent liquid becomes invisible because when refractive index of the material of the lens is equal to the refractive index of the liquid in which lens is placed under this condition no bending of light takes place when it travels from liquid to the lens, so both will start behaving like both are same things.
Explanation:
hope it helps :))
Answer:
Total distance, ![d=\dfrac{xyt}{(x+y)}](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7Bxyt%7D%7B%28x%2By%29%7D)
Explanation:
It is given that,
Speed of Aaron from home is y mph and walk back at x mph. Let t is the total time he spend in walking and jogging. Let d is the distance covered.
We he moves from home to destination, time is equal to, ![\dfrac{d}{x}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bx%7D)
Similarly, when he move back to home, time taken is equal to ![\dfrac{d}{y}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7By%7D)
Total time taken is equal to :
![\dfrac{d}{x}+\dfrac{d}{y}=t](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bx%7D%2B%5Cdfrac%7Bd%7D%7By%7D%3Dt)
![d(\dfrac{1}{x}+\dfrac{1}{y})=t](https://tex.z-dn.net/?f=d%28%5Cdfrac%7B1%7D%7Bx%7D%2B%5Cdfrac%7B1%7D%7By%7D%29%3Dt)
![d=\dfrac{t}{(\dfrac{1}{x}+\dfrac{1}{y})}](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7Bt%7D%7B%28%5Cdfrac%7B1%7D%7Bx%7D%2B%5Cdfrac%7B1%7D%7By%7D%29%7D)
![d=\dfrac{xyt}{(x+y)}](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7Bxyt%7D%7B%28x%2By%29%7D)
So, the distance he speed in walking and jogging is
. Hence, this is the required solution.
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Adoption process of the SI system by all the worlds nations is called Metrication or metrification.
Metrication or metrification is defined as conversion to the metric system of units of measurement. Earlier, many traditional and local systems were used for conversion but since last 2 centuries, most of the countries have adopted the SI system that is widely used as system of measurement.
Answer:
a = -0.05 m/s² (negative sign shows deceleration)
Explanation:
In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:
2as = Vf² - Vi²
where,
a = minimum acceleration required = ?
s = minimum distance covered = 250 m
Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)
Vi = Initial Velocity = 5 m/s
Therefore,
2a(250 m) = (0 m/s)² - (5 m/s)²
a = - (25 m²/s²)/(500 m)
<u>a = -0.05 m/s²</u> (negative sign shows deceleration)