Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of expansion twice as large as the second bar. How does the original volume of the first bar compare to the original volume of the second bar

The original volume of the first bar is half of the original volume of the second bar.

Explanation:

The coefficient of cubic expansivity of substances is given by;

γ = ΔV ÷ (Δθ)

Given: two metal bars with equal change in volume, equal change in temperature.

Let the volume of the first metal bar be represented by , and that of the second by .

Since they have equal change in volume,

Δ  = Δ = ΔV

For the first metal bar,

2γ = ΔV ÷ (Δθ)

⇒    Δθ =  ΔV ÷ (2γ)

For the second metal bar,

γ = ΔV ÷ (Δθ)

⇒  Δθ = ΔV ÷ (γ)

Since they have equal change in temperature,

Δθ of first bar = Δθ of the second bar

ΔV ÷ (2γ)     =   ΔV ÷ (γ)

So that;

(1 ÷ 2) = (1 ÷ )

2 =  

=

Thus, original volume of the first bar is half of the original volume of the second bar.

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

the answer is b, have negligible mass