Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as

where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,

And since the sign is positive, the direction is the same as the velocity (south).
Answer:
D
Explanation:
As we progress and learn more, scientists make new discoveries that can contradict earlier discoveries, and since are technology is better today we are able to discover a lot more to add to or change our previous theories.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
Compare the initial mass to the final mass.