Answer:
1.37 rad/s
Explanation:
Given:
Total length of the tape is,
m
Total time of run is,
hours
We know, 1 hour = 3600 s
So, 2.1 hours = 2.1 × 3600 = 7560 s
So, total time of run is,
s
Inner radius is, 
Outer radius is, 
Now, linear speed of the tape is, 
Let the same angular speed be
.
Now, average radius of the reel is given as the sum of the two radii divided by 2.
So, average radius is, 
Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

Therefore, the common angular speed of the reels is 1.37 rad/s.
No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.
Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !
Answer:
harmful effects
1. that will cause air pollution
2. that will destroy our earth
If an object's speed changes, or if it changes the direction it's moving in,
then there must be forces acting on it. There is no other way for any of
these things to happen.
Once in a while, there may be <em><u>a group</u></em> of forces (two or more) acting on
an object, and the group of forces may turn out to be "balanced". When
that happens, the object's speed will remain constant, and ... if the speed
is not zero ... it will continue moving in a straight line. In that case, it's not
possible to tell by looking at it whether there are any forces acting on it.
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