The work that is required to increase the speed to 16 knots is 14,176.47 Joules
If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;
5.44×10^3 kg = 12 knots
For an increased speed to 16knots, we will have:
x = 16knots
Divide both expressions
To get the required work done, we will divide the mass by the speed of one knot to have:
Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules
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I belive it would power I’m not 100% sure. If it’s not power then force
Refer to the figure shown below, which is based on the given figure.
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
![4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0](https://tex.z-dn.net/?f=4.9%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%20%5D%5E%7B2%7D%20-%20v_%7B0%7D%20sin%28%5Ctheta%20-%20%20%5Calpha%20%29%20%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%5D%20-%20X%20sin%20%5Calpha%20%20%3D%200)
Hence obtain
![aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha](https://tex.z-dn.net/?f=aX%5E%7B2%7D-bX%3D0%20%5C%5C%20where%20%5C%5C%20a%3D4.9%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%5D%5E%7B2%7D%20%5C%5C%20%20b%20%3D%20cos%20%5Calpha%20%5C%2C%20%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%20%2B%20sin%20%5Calpha%20)
The non-triial solution for X is
Answer:
![X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}](https://tex.z-dn.net/?f=X%3D%20%5Cfrac%7Bsin%20%5Calpha%20%20%2B%20cos%20%5Calpha%20%20%5C%2C%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%7B4.9%20%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20%5C%2C%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%20%20%5D%5E%7B2%7D%7D%20)
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m
The kind of cosmic accident which is believed to form many irregular shaped galaxies is known as Galactic Collisions. Galactic Collisions are cosmic accidents that create irregularly shaped galaxies. The Andromeda-Milky Way Collision is one example of galactic collision that is predicted to occur in the next four (4) billion years, where the Andromeda and the Milky Way galaxy collide will collide with each other to form a new galaxy.
light speed in vacuum = 3.8 * 10^8
Distance (Given) = 3.5 ft
now, time = distance/speed = 3.50 / 3.8 * 10^8 = 9.21 * 10^-9 s = 9.21 nanoseconds
So, your answer is 9.21 nano-seconds...
