Amodel for the density δ of the earth’s atmosphere near its surface isδ=619.09−0.97pwhere p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter. if we take the surface of the earth to be a sphere with radius 6370 km, then this model is a reasonable one for: 6.370×106≤p≤6.375×106 use this model to estimate the mass of the atmosphere between the ground and an altitude of 5 km.

3.751*10^18kg

Step-by-step explanation:

δ =619.09−0.000097pequa1 where p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter.

Calculating the density of air at 5km above earth surface

P = 5000m + 6370000m = 6.375*10^6m

δ = 619.09 -(.000097* 6.375*10^6)

δ = 0.715kg/m^3 = density

Since Mass = density*volume...equ2

To calculate volume of air around the spherical earth at height 5km

V = (4/3 pai R^3) - (4/3pai r^3) ...equation 3 where R =6.375*10^6m, r = 6.37*10^6

Substituting R and r in equation 2 to solve for volume of air

V = 1.085*10^21 - 1.08*10^21

V = 5.25*10^18m^3

Substituting δ and V into equation 2 to solve for mass of air

M = 0.715 * (5.25*10^18)

M = 3.751*10^18kg

step-by-step explanation:

a cube has 6 sides

the probability of getting 5 on the first cube is 1/6 since only one side of the cube has 5 on it.

there are 3 sides of the cube with even number on it [2,4,6]

therefore the probability of getting even number is 3/6

for both events to happen together, the probability would be

1/6 x 3/6 = 12/36

which simplifies to 1/12

step-by-step explanation:

f(gx)=f(1/x+2)

=(1/x+2)+2

=(1+2x)/2+2

=(1+2x+4)/2

= (2x+5)/2