Given that the mass of an average linebacker at ursinus college is 250 lbs and the radius of a pea is 0.50 cm, calculate the number of linebackers that would be required to be stuffed into the volume of a pea in order to obtain the same density as an alpha particle.

1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle

Step-by-step explanation:

Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is

V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³

the mass in that volume would be  m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)

The density of an alpha particle is  ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be

ρ= m/V

since both should be equal ρ=ρa , then

ρa= m/V =N*L/V → N =ρa*V/L

replacing values

N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg  = 1.544*10⁹ Linebackers

N=1.544*10⁹ Linebackers

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8