Question

question 27
1 pts
given the equation for density is d = m/v, what happens to the density of the air when the gas in a balloon is heated and the volume of the balloon increases?
the density goes up and the balloon rises
the density goes down and the balloon sinks
the density goes down and the balloon rises
the density goes up and the balloon sinks
question 28
1 pts
solve the following equation for the change in temperature, δt. q = mcδt
δt = qc/m
δt = m/qc
δt = q/mc
δt = mq/c

Explanation:

Step1: Analyze density - volume relationship

Given density formula $d=\frac{m}{V}$. When the gas in the balloon is heated and volume $V$ increases while mass $m$ remains constant (assuming no gas escapes).

Step2: Determine density change

As $V$ increases and $m$ is constant, the value of $\frac{m}{V}$ (density $d$) decreases. A less - dense object in a denser medium (surrounding air) will rise. So the balloon rises.

Step3: Solve for $\Delta T$ in $q = mc\Delta T$

Divide both sides of the equation $q = mc\Delta T$ by $mc$ to isolate $\Delta T$. We get $\Delta T=\frac{q}{mc}$.

Answer:

Question 27: the density goes down and the balloon rises
Question 28: $\Delta T=\frac{q}{mc}$