Answer:
253 years
Explanation:
a = Distance in Astronomical units from the sun = 40 AU
P = Orbital period in years
From Kepler's law we have
The orbital period of the Kuiper belt object is 253 years.
A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the coefficient of static friction is .65. What is
the force of static friction?
1 answer:
m = mass of the car = 1800 kg
g = acceleration due to gravity = 9.8 m/s²
= force of gravity on the car
force of gravity on the car is given as
= mg
inserting the values
= (1800) (9.8) = 17640 N
= static frictional force acting on the car
from the force diagram of the car , the static frictional force is opposite to the parallel component of force of gravity on the car . hence static frictional force mus balance the component of force of gravity parallel to incline surface.
hence
=
Sin7
= (17640) Sin7
= 2149.8 N
You might be interested in
The derivative of the function space as a function of time is equal to a function of speed as a function of time.
The velocity vector is given by the vector sum of the velocities of both axes.
If you notice any mistake in my english, please let me know, because I am not native.
The velocity vector is given by the vector sum of the velocities of both axes.
If you notice any mistake in my english, please let me know, because I am not native.
Answer:
The magnetic field strength needed is 1.619 T
Explanation:
Given;
Number of turns, N = 485-turn
Radius of coil, r = 0.130 m
time of revolution, t = 4.17 ms = 0.00417 s
average induced emf, V = 10,000 V.
Average induced emf is given as;
V = -ΔФ/Δt
where;
ΔФ is change in flux
Δt is change in time
ΔФ
where;
N is the number of turns
B is the magnetic field strength
A is the area of the coil = πr²
θ is the angle of inclination of the coil and the magnetic field,
V = NBACos0/t
V = NBA/t
B = (Vt)/NA
B = (10,000 x 0.00417) / (485 x π x 0.13²)
B =1.619 T
Thus, the magnetic field strength needed is 1.619 T