Answer:
I.D.K but the same measure they use to judge will be used to judge them
Explanation:
Answer:
the impedance of the circuit is 25.7 ohms.
Explanation:
It is given that,
Voltage, V = 50 volts
Frequency, f = 60 Hz
Resistance, R = 25 ohms
Capacitive resistance, 
Inductive resistance, 
We need to find the impedance of the circuit. It is given by :
Z = 25.7 ohms
So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.
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:
350 N/m
Explanation:
If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m
k = 70 / 0.2 = 350 N/m
The spring constant is 350 N/m
