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:
A. 2086 yd²
Step-by-step explanation:
SA = 2lw + 2wh + 2lh
SA = 2*17*20 + 2*20*19 + 2*17*19
SA = 680 + 760 + 646
SA = 2086
Hope this helps :)
First simplify the square roots:

Then simplify the last two terms:

Since 61 is prime, you can't take a rational root out of it.
Answer:
z' (-9,2)
Step-by-step explanation:
I recomend using desmos to show the relashinship
If you don't want to, thwn do this
Preimage
Since we are finding z, we only use the final cordinates
z'=(-9,2)