Your answer for this question is the third option.
Answer:
38.8 m/s
Explanation:
Force F(x) = 6 - 2x + 6x²
work


W = mv²/2=7v²/2 = 3.5v² = 5261
v = 38.8 m/s
Answer:
v₁ = 4 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]

where:
m₁ = mass of the tank = 500 [kg]
v₁ = velocity of the tank after firing the missile [m/s]
m₂ = mass of the missile = 20 [kg]
v₂ = velocity of the missile after firing = 100 [m/s]
![(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]](https://tex.z-dn.net/?f=%28500%2Av_%7B1%7D%29%3D%2820%2A100%29%5C%5Cv_%7B1%7D%3D2000%2F500%5C%5Cv_%7B1%7D%3D4%5Bm%2Fs%5D)
The answer is <span>b. False. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h. Give it a try man
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Answer:
R = 3.88 m
Explanation:
As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:
V₀ = Launch Speed = 6.7 m/s
θ = Launch Angle = 29°
R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?
The value of the range of a projectile is given by the following formula:
R = (V₀² Sin 2θ)/g
R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]
R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]
<u>R = 3.88 m</u>