As simply put as I believe is possible, it's the immediate space-time co-oridinate in which the observer is not in motion. All measurements by the observer are in relation to that 'place'
Answer:
(a) The "angular speed" is 5.88 rad/s.
Explanation:
Given values,
The length of the bar is L = 2m
The weight of the bar is w = 90 N
The metal bar is hanging vertically from the ceiling by a frictionless pivot
The mass of the ball is m = 3kg
The distance between the ceiling and the ball is d = 1.5m
(a) Calculating the angular speed:
The angular speed is 5.88 rad/s.
(b) The "angular momentum" is conserved because the torque is not exerted by "the pivot" on the system about the "axis of rotation" but the "linear momentum" is not conserved because "the pivot" exerts a "vertical" and a "horizontal force" on the system during the collision.
Heat Transfer. Heat transfer is the exchange of thermal energy between physical systems.
The given configuration is : [Ar]3d²
The inner shells of this species has the noble gas, Argon (Ar) configuration while its valence configuration is 3d².
Now the atomic number of Argon = 18
Therefore, the species will have an atomic number = 18 + 2 = 20
Based on the periodic table the species with atomic number 20 is calcium
Ca = [Ar] 3d²
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<span>v(t)= -gt - ve(ln((m - rt)/m))
x(t) = ∫v(t) dt
= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)
Let u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du
So x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c
= mve/r [ (m - rt)/m * (ln((m - rt)/m) - 1] - ½gt²
= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²
If g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s
x(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²
≈ 138390 m
= 138.39km</span>