Answer:
You already wrote the equation
Step-by-step explanation:
Let its width be w and length= w+6
2(w+6+w)=52
4w+12=52
4w=40
w=10
Area=10*(10+6)=160
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
All you simply have to do is multiply 20 x 0.35. Multiplying it by 0.35 will turn it into the number that makes up 35%. The answer would be they made 7 shots.
Your answer would be 108 surface area
