At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t
2 answers:
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Answer:
A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)
B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist
C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf
Say mass of husband is m1
Mass of the wife is m2
Velocity of the husband is v1
Velocity of the wife is v2
According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf
The momentum equation is
m1v1+m2v2= (m1+m2)vf
D. To solve for vf we need to make it subject of formula
vf= {(m1v1) +(m2v2)}/(m1+m2)
E. Substituting our given data
vf=
{(1570*58)+(2550*54)}/(1570+2558)
vf=91060+137700/4120
vf=228760/4120
vf=55.52m/s
Their speed after collision is 55.52m/s
Answer:
The value of the average convection coefficient is 20 W/Km².
Explanation:
Given that,
For first object,
Characteristic length = 0.5 m
Surface temperature = 400 K
Atmospheric temperature = 300 K
Velocity = 25 m/s
Air velocity = 5 m/s
Characteristic length of second object = 2.5 m
We have same shape and density of both objects so the reynold number will be same,
We need to calculate the value of the average convection coefficient
Using formula of reynold number for both objects
Here,
Put the value into the formula
Hence, The value of the average convection coefficient is 20 W/Km².