Answer:
note duration are measured in beats
(a) 3675 N
Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

where
is the horizontal component of the force
m is the mass of the passenger
is the horizontal component of the acceleration
Here we have
m = 75.0 kg

Substituting,

(b) 3748 N, 11.3 degrees above horizontal
In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

where we used
as acceleration of gravity.
So, this is the vertical component of the force exerted by the seat on the passenger:

and therefore the magnitude of the net force is

And the direction is given by

When its tangential speed is constant
<span>Although the speed of an object that has a uniform circular motion is constant, its velocity is </span>not constant<span>. Not only that, but it is actually changing constantly.</span><span>
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Answer: im just trying to get some points so i can answer my question.
sorry
Explanation:
Explanation:
Eq 2 ) is Equation for force on a charge Q1 due to charge Q2 separated by a distance R .. it is Equation for Coloumb's law !!
Where , k = 1/4π€
Eq 3 ) is Equation for force on mass m1 due to mass m2 separated by distance R .. it is Newton's law of Gravitation
Where , G = universal Gravitation constant