A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H
1 answer:
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Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components
Answer:
,
Explanation:
The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.
Therefore, we can write:
where
is the thrust force generated by each engine of the jet
is the drag force
Solving for Fd,
The velocity of the jet is
So, the rate at which the drag force does work (which is the power) is
and substituting
we find
Converting into horsepower,
Answer:
B). to the right
Explanation:
Since the direction of magnetic field is into the page
So here we know that
now the velocity is from bottom to top
so we have
now the force on the moving charge is given as
now we have
so force will be towards Right