Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
The air inside becomes warmer, and that makes it LESS DENSE.
Less-dense air FLOATS in denser air. Since it's inside the balloon, it picks the balloon up when it floats.
Answer:48 V
Explanation:
Given
Three charged particle with charge
Electric Potential is given by
Distance of 
from 
similarly 
Potential at is
![V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]](https://tex.z-dn.net/?f=V_%7Bnet%7D%3Dk%5B%5Cfrac%7Bq_1%7D%7Bd_1%7D%2B%5Cfrac%7Bq_2%7D%7Bd_2%7D%2B%5Cfrac%7Bq_3%7D%7Bd_3%7D%5D)
![V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}](https://tex.z-dn.net/?f=V_%7Bnet%7D%3D9%5Ctimes%2010%5E9%5B%5Cfrac%7B50%7D%7B10%7D-%5Cfrac%7B80%7D%7B12%7D%2B%5Cfrac%7B70%7D%7B10%7D%5D%5Ctimes%2010%5E%7B-9%7D)
#1). Anthony does the same amount of work as Angel, with <em>more power</em>.
#2). Power = (Work)/(Time) = 41,000 J / 500 s = <em>82 watts .</em>
#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>
Answer:
The baseball will stay in motion until another force act upon it.
Explanation:
