Answer: Change in momentum=9.4kgm/s
Impulse=9.4kgm/s
Explanation:
Change in momentum=5.1-(-4.3)=5.1+4.3=9.4kgm/s
Impulse=Change in momentum
There impulse=9.4kgm/s
In emission nebulae, there are interstellar clouds of hydrogen, which glow red because of the intense radiation of hot stars inside the nebula
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
Answer:
de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
Explanation:
Given;
mass of bullet, m = 28 g = 0.028 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
λ = h / mv
where;
λ is de Broglie wavelength of the bullet
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ = h / mv
λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)
λ = 3.093 x 10⁻³⁵ m
Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
