Answer:
Explanation:
This is an application of Newton's second Law.
Formula
F = m * a
F = 300 N
m = 100 kg
a = ?
F = m * a
300N = 100 kg * a Divide by 100
300N/100kg = a
a = 3 m/sec^2
Answer:
The answer is the principal Quantum number (n)
Explanation:
The principal quantum number is one of the four quantum numbers associated with an atom.
It is denoted by a number n=1,2,3,4 etc
It tells both size (directly) and energy (indirectly) of an orbital.
When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.
It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.
Answer:
Explanation:
I suppose it has to do with the way the diagram is drawn. The heat does not reflect which makes both A and B incorrect.
C would have nothing to do with either reflection or refraction.
That only leaves D which is the answer.
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet
b) tangential speed
v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude
a = 3.62 x 10⁻³ m/s²
