Answer:
ok send the work so that I can do it
Answer:
Explanation:
At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation
Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.
Simplifying
v = 
Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get
v = ![[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B2%5Ctimes%20%281.2%5Ctimes10%5E%7B-5%7D%29%5E2%282182-1.225%29%7D%7B9%5Ctimes%201.8%5Ctimes10%5E%7B-5%7D%7D)
[/tex]
v = 387 x 10⁻⁵ m/s
Terminal velocity = 387 x 10⁻⁵ m/s
Time taken to fall a distance of 100 m
= 
= 2.6 x 10⁴ s.
In the conservation of mass, mass is never created or destroyed in chemical reactions in the same way water is not created or destroyed it is only transferred from one form to another and its mass is always conserved.
Answer:
B and C
Explanation:
Because EMWs are varying magnetic and electric radiation traveling at 90° to each other propagating energy form one place to another through vibration of these magnetic and electric fields
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
