A)they can interbreed and produce fertile offsprings
Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Missing part of the question: determine the magnitude of Porsha's acceleration.
Given the data in the question;
- Mass of Porsha;

- Mass of Zorn;

- Force of Porsha push;

Magnitude of Porsha's acceleration; 
To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

Where m is the mass of the object and a is the acceleration.
We substitute the mass of Porsha and the force he used into the equation
Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Learn more: brainly.com/question/25125444
Answer:
B
Explanation:
if no force acts upon it it will just continue moving
hope this helps!! have a wonderfull day!
I think it is B, because the sun’s size is pretty average
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁ ![ln[\frac{P_{1} }{P_{2} }]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7BP_%7B1%7D%20%7D%7BP_%7B2%7D%20%7D%5D)
= <em>mRT</em>₁
[∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039) ![ln[\frac{2.0}{1.0}]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7B2.0%7D%7B1.0%7D%5D)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> = 
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ