Each
ion contains three extra protons. Hence, the extra charge on each
=
C
Total charge = 0.035 pC
Total charge (Q) =
C
Let the number of
ions be n.
According to question:



n = 72917
Hence, the total number of ions needed to be transferred is 72917
The answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg
Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s