Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
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Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
Where
and the position vector
using the determinant method to expand the cross product in order to determine the torque we have
by expanding we arrive at
since we have determine the vector value of the toque, we now compare with the torque value given in the question
if we directly compare the j coordinate we have
Explanation:
The given data is as follows.
m = 5000 kg, h = 800 km =
, r = R + h =
kg, G =
As we know that,
v =
And, it is known that formula to calculate angular velocity is as follows.
v =
=
=
Thus, we can conclude that speed of the satellite is .