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
6x - 9x - 5 =
combine like terms
6 - 9 = - 3
- 3x - 5 =
leave x by itself
- 3x - 5 =
+3x
___
0
- 5 = 3x
solve for x (division)
- 5 = 3x
___
3
-5/3 = x OR x = -5/3
The answer to this equation is 5.
Answer:
fraction 4/12 or 1/3
decimal .333333333
percent 33%.333333
Step-by-step explanation:
The answer is: 1.49
i used a calculator.
