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
<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365
Fsf = Static frictional force = (coefficient of static friction) * (Normal force)
So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
</span>
Answer:
For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.
Explanation:
Hope this helps
Answer:
A) 0 miles
Explanation:
The displacement is the distance between the starting point and the end point of the route.
In this case, even if Matino takes a whole tour around the city, since<u> it ends in the same place where it started</u>, the difference between the starting and finishing point is zero, so its total displacement is zero.
Care must be taken to distinguish the terms of displacement and distance traveled, beacause they are not the same, since in this case the distance traveled would be 3.15miles, but the displacement is zero, because it ends at the point where it started.
