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
Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.
Answer:
he will use kinetic energy to break through the door
Explanation:
because kinetic energy is movement and force
I’m sorry I don’t understand this language
Answer:
the location of the center of gravity for the entire body is 1.08 m
Explanation:
Given the data in the question;
w1 = 458 N, y1 = 1.34 m
w2 = 120 N, y2 = 0.766 m
w3 = 89.8 N, y2 = 0.204 m
The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.
so,
= (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )
so we substitute in our values
= (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )
= 723.9592 / 667.8
= 1.08 m
Therefore, the location of the center of gravity for the entire body is 1.08 m
