Answer:
Explanation:
The sum of the pore along the plane is expressed according to Newton's law
Fn-Ff = ma
Fn is the moving force
Ff = nR = frictional force
m is the Mass
a is the acceleration
Substitute the given values
Fn - nR = ma
Fn - tan31°(mgcostheta) =3.9(9.8)
Fn - tan31(3.9(9.8)cos31) = 3.9(9.8)
Fn - tan31(38.22cos31)= 38.22
Fn - 32.76tan31 = 38.22
Fn-19.68 = 38.22
Fn = 38.22+19.68.
Fn = 57.90N
Hence Fn (moving force) of the inclined block is 57.90
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
IMA stands for ideal mechanical advantage, which is the theoretical force amplification factor on an ideal mechanical device free of friction, deformations, etc.
If the applied force (effort) is 50N, then the force applied to the resistance is multiplied by the IMA=2 to get 100N.
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m