C.0 because it didn't move<span />
Answer:
F = 2389.603 N
Explanation:
Given:
Mass m = 1,369.4 kg
Initial velocity u = 28.9 m/s
Final velocity v = 20 m/s
Time t = 5.1 s
Find:
Net force
Computation:
a = (v - u)/t
a = (20 - 28.9)/5.1
a = -1.745 m/s²
F = ma
F = (1369.4)(1.745)
F = 2389.603 N
Answer:
The law of conservation of momentum states that the total momentum of interacting objects does not <u>change</u>. This means the total momentum <u>before</u><u> </u>a collision or explosion is equal to the total momentum <u>after</u><u> </u>a collision or explosion.
Answer:
B. No. He presented no scientific data to support his claim.
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
