Asteel sphere sits on top of an aluminum ring. the steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at 0 degrees celsius. the aluminum ring ( a = 2.4 x 10^-5/degrees celsius) has an inside diamter of 3.994 cm at 0 degrees celsius. at what temperature will teh sphere fall through teh ring? ? mulitple choice questiona, 461.5 degrees celsiusb. 208.0 degrees celsiusc. 115.7 degrees celsiusd. 57.7 degrees celsius

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

+15-35=-20

going in opposite directions

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000