Answer:
because they may not actually be moving be just percieved as so
Explanation:
Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
Answer:
The value is 
Explanation:
From the question we are told that
The first amplitude of the wave is 
The first depth is 
The second amplitude is 
The second depth is 
Generally from the spatial wave equation we have
=> 
So considering the ratio of the equation for the two depth
=> 
=> 
=>
Answer:
a= -0.83m\s^2
Explanation:
a = v \ t
a = -25 \ 30 = -0.833 m\s^2
the object is slowing down 0.83 meter every second
