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:
18, 19, 20, 21
Step-by-step explanation:
Just like any of these problems, we should start by forming an equation for us to get a reference and plug in. We'll be using x as our variables.
As we have four consecutive integers (and not multiples) we can assume that the integers will be x, x+1, x+2, and x+3.
The sum of the two largest integers we have equals: n+2 + n+3 = 2n+5
and three times the sum of the two smallest = 3(n + n+1) = 6n+3
and the sum of t he two largest subtracted from three times the sum of the two smallest = (6n+3) - (2n+5) = 4n-2
4n-2=70
4n=68
n = 18.
n = 18, n+1 = 19, n+2 = 20, n+3 = 21
Answer:
Cos A = 24/25 & Tan A = 7/24
Step-by-step explanation:
See the attached figure.
Construct a right triangle and sign one angle as A
So, Sin A = opposite/hypotenuse = 7/25
∴hypotenuse = 25 and opposite = 7
∴ adjacent = √(25² - 7² ) = √576 = 24
∴ Cos A = adjacent/hypotenuse = 24/25
And
Tan A = opposite/adjacent = 7/24
The greatest common factor is 8
