Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.
This problem is to let you practice using Newton's second law of motion:
Force = (mass) x (acceleration)
-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.
-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.
-- Now you know ' F ' and ' mass '. Use Newton's second law of motion to calculate the plane's acceleration.
Current flows from High Potential (Positive) to Low potential (Negative)
So, option D is your answer!
Hope this helps!
Answer:
E3 = 3.03 10⁻¹⁶ kJ, E4 = 4.09 10⁻¹⁶ kJ and E5 = 4.58 10⁻¹⁶ kJ
Explanation:
They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units
n = 3 λ = 656.3 nm = 656.3 10⁻⁹ m
n = 4 λ = 486.1 nm = 486.1 10⁻⁹ m
n = 5 λ=434.0 nm = 434.0 10⁻⁹ m
Let's use the Planck equation
E = h f
The speed of light equation
c = λ f
replace
E = h c /λ
Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s
Let's calculate the energies
E = 6.63 10⁻³⁴ 3 10⁸ / λ
E = 19.89 10⁻²⁶ /λ
n = 3
E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹
E3 = 3.03 10⁻¹⁹ J
1 kJ = 10³ J
E3 = 3.03 10⁻¹⁶ kJ
n = 4
E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹
E4 = 4.09 10⁻¹⁹ J
E4 = 4.09 10⁻¹⁶ kJ
n = 5
E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹
E5 = 4.58 10⁻¹⁹ J
E5 = 4.58 10⁻¹⁶ kJ
Answer:
5
Explanation:
The d subshell has 5 orbitals, each capable of holding a maximum of two electrons. Hund's rule tells us that every orbital in a sub-level must first be singly occupied by electrons before any orbital is doubly occupied. Therefore five electrons will fill the five orbitals within the d subshell.
