I learned the equation as P•V=k•T .
So x=1, y=1, and z= -1 .
His position x = V* t = 6.7*56= 375.2 m from point a
Hi there!
We can use the following equation for a point charge in a magnetic field:

= Force due to magnetic field (7.6 × 10⁻¹³N)
= Charge of particle (1.6 × 10⁻¹⁹ C)
= velocity of particle (6.6 × 10⁶ m/s)
= Magnetic field strength (1.8 T)
Or, without the cross product:

θ = angle between particle's velocity and field
We can rearrange to solve for theta:

Solve for theta:

Explanation:
(a) The given data is as follows.
Length of the rod, L = 0.83 m
Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)
At the lowest point, angular speed of the rod (
) = 5.71 rad/s
First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.
I = 
=
= 
= 0.00631 + 0.415
= 0.42131 
Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.
K = 
= 
= 6.86 J
Hence, kinetic energy of the rod at its lowest point is 6.86 J.
(b) According to the conservation of total mechanical energy of the rod, we have


or, mgh = K = 6.86 J
Therefore, h =
= 
= 0.584 m
Hence, the center of mass rises 0.584 m far above that position.
Answer: 22.01 rpm.
Explanation:
If we assume no external torques are present, the angular momentum must be conserved.
L1 = L2
I1 . ω1 = I2 ω2 (1)
I1 = MR2 /2 + m1. R2 + m2. R2 + m3. R2
I2 = MR2 /2 + m1. R2 + m3. R2
(as the 28.0 kg child moves to the center, he has no part anymore in the rotational inertia)
As I units are SI units, it is advisable to convert the angular velocity to rad/sec, as follows:
22.0 rev/min. (2π/rev) . (1min/60 sec) = 2.3 rad/sec
Solving for ω2 in (1):
ω2 = 2.31 rad/sec = 2.31 (1 rev/2 π). (60sec/1min) = 22.01 rpm