velocity of the physics instructor with respect to bus

acceleration of the bus is given as

acceleration of instructor with respect to bus is given as

now the maximum distance that instructor will move with respect to bus is given as




so the position of the instructor with respect to door is exceed by

so it will be moved maximum by 3 m distance
The answer is C, individuals copy works to view at a later time.
Answer:
The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.
Explanation:
According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.
σx . σpx ≥ h/4π
where,
h is the Planck´s constant
If σx = 5 × 10⁻¹²m,
5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π
σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹
Answer:
4.535 N.m
Explanation:
To solve this question, we're going to use the formula for moment of inertia
I = mL²/12
Where
I = moment of inertia
m = mass of the ladder, 7.98 kg
L = length of the ladder, 4.15 m
On solving we have
I = 7.98 * (4.15)² / 12
I = (7.98 * 17.2225) / 12
I = 137.44 / 12
I = 11.45 kg·m²
That is the moment of inertia about the center.
Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that
τ = 11.453 kg·m² * 0.395 rad/s²
τ = 4.535 N·m