D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.
Answer:
18.63 N
Explanation:
Assuming that the sum of torques are equal
Στ = Iα
First wheel
Στ = 5 * 0.51 = 3 * (0.51)² * α
On making α subject of formula, we have
α = 2.55 / 0.7803
α = 3.27
If we make the α of each one equal to each other so that
5 / (3 * 0.51) = F2 / (3 * 1.9)
solve for F2 by making F2 the subject of the formula, we have
F2 = (3 * 1.9 * 5) / (3 * 0.51)
F2 = 28.5 / 1.53
F2 = 18.63 N
Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
By definition, the potential energy is:
U = qV
Where,
q: load
V: voltage.
Then, the kinetic energy is:
K = mv ^ 2/2
Where,
m: mass
v: speed.
As the power energy is converted into kinetic energy, we have then:
U = K
Equating equations:
qV = mv ^ 2/2
From here, we clear the speed:
v = root (2qV / m)
Substituting values we have:
v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
v = 3.56 × 10 ^ 7 m / s
Then, the centripetal force is:
Fc = Fm
mv ^ 2 / r = qvB
By clearing the magnetic field we have:
B = mv / qr
Substituting values:
B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
B = 3.43 × 10 ^ -3 T
Answer:
A magnetic field that must be experienced by the electron is:
B = 3.43 × 10 ^ -3 T
