hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
Answer:
Both of them.
Explanation:
They are both because when your analyzing data , that is what happen's.
Answer:
Explanation:
The spaceship has traveled 3% of the distance to a space station and it has traveled 
miles.
Let the total distance from the ship's starting point to the space station be x.
This means that:
The total distance to be traveled is 
.
Therefore, the distance left to travel is:
Answer:
The correct option is D
Explanation:
From the question we are told that
The refractive index of oil film is 
The thickness is 
Generally for constructive interference
![2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}](https://tex.z-dn.net/?f=2t%20%20%3D%20%20%5Bm%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%2A%20%20%5Cfrac%7B%5Clambda%7D%7Bk%7D)
For reflection of a bright fringe m = 1
=> ![2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}](https://tex.z-dn.net/?f=2%20%2A%20%28290%2A10%5E%7B-9%7D%29%20%20%3D%20%20%5B1%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%2A%20%20%5Cfrac%7B%5Clambda%7D%7B1.48%7D)
=> 
This wavelength fall in the range of a yellow light
