<span>A) x = 41t
The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So
H = 120*cos(70) = 120*0.34202 = 41.04242
So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.</span>
Answer:
0.29 m/s due west.
Explanation:
According to newton's second law,
Net force acting on an object = mass×acceleration
From the question,
F+F₁+F₂ = ma................ Equation 1
Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.
Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg
substitute into equation 1
4080-680-1160 = 7660(a)
2240 = 7660a
Therefore,
a = 2440/7660
a = 0.29 m/s due west.
Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 =
Answer: It wouldn't be as modern as today is we would be back to using oil and other things from back then there wouldn't be cars everything would be less machined.
