Answer:
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
Answer:
Step-by-step explanation:
The volume of a cube is
V=s^3, we are told the volume is 10u^3 and that it will be filled with cubes having a side of 1/2 so
n(1/2)^3=10, where n will be the number of these small cubes
n(1/8)=10 upon multiplying each side by 8
n=80
So it will take 80 cubes of side length 1/2 to have a volume equal to 10u^3
Do 6x 266 in your calculator that’s ur answer
Answer:
the answer will be 129 i think
