Answer:The human eye is sensitive to yellow-green light having a frequency of about 5.5*10^{14} ... What is the energy in joules of the photons associated with this light? ... As the wavelength and frequency of a wave are related, we can find the energy ... In order to find this value, we need Planck's Constant, h=6.626×10−34 J⋅s h ...
Explanation:
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height

To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height



So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point



Solving the quadratic equation



So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:


Since the projectile is going down the velocity at the instant it reaches the ground is:
