Explanation:
<u>Formula:</u>
<u>d = distance given</u>
<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>
<u>Substitute the given values into the formula for velocity</u><u>:</u>
velocity is shortened for v.
8 (distance) divided by 4 (time) equals the velocity.
<u>Solve:</u>
The velocity of the toy car equals: B. 2 m/s.
They all have a special way in their age how they want to get taught things but there abilities make them special
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
Answer:
A 300-kg crate is placed on a adjustable incline plane. as one end of the incline is raised, the crate begins to move downward. if the crate slides down the plane with the acceleration of .70 m/s2 , when the incline is 25 degrees, what is the coefficient of kinetic friction between ramp and crate?
Explanation:
