Brownian motion<span> or pedesis is the </span>random motion<span> of particles suspended in a fluid </span>
Answer: TheTroposphere contains 80% of the total gas in the atmosphere
Answer:
10.89 J.
Explanation:
The following data were obtained from the question:
Mass (m) = 12.5 kg
Velocity (v) = 1.32 m/s
Work done =?
To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:
Mass (m) = 12.5 kg
Velocity (v) = 1.32 m/s
Kinetic energy (K.E) =?
K.E = ½mv²
K.E = ½ × 12.5 × 1.32²
K.E = 6.25 × 1.7424
K.E = 10.89 J
The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.
Answer:
Energy = 1.38*10^13 J/mol
Explanation:
Total number of proton in F-19 = 9
Total number of neutron in F-19 = 10
Expected Mass of F-19
= 9*1.007 + 10*1.008 = 19.152 u
Actual mass of F-19 = 18.998 u
Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)
= 143.234 MeV
Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23
= 1.38*10^13 J/mol
Answer:
Explanation:
For elestic collision
v₁ = 
![v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}](https://tex.z-dn.net/?f=v_2%20%3D%20%5Btex%5D%5Cfrac%7B%28m_2-m_1%29u_2%7D%7Bm_1%2Bm_2%7D%20%2B%5Cfrac%7B2m_1u_1%7D%7Bm_1%2Bm_2%7D)
[/tex]
Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg , v₁ and v₂ are velocity of goalie and puck after the collision.
v₁ = 0 + ( 2 x .15 x22 )/ 77.15
= .085 m / s
Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.
v₂ = (.15 - 77)x 22 / 77.15 +0
= - 21.91 m /s
=Velocity of puck will be - 21.91 m /s in the direction opposite to original velocity of ball before collision.
