Answer:
D.-4.798m/s
Explanation:
Greetings !
Given values

Solve for V of the given expression
Firstly, recall the velocity-time equation

plug in known values to the equation

solve for final velocity

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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
Answer:
Input force of pulley system = 200 N
Explanation:
Given:
Mechanical advantage of pulley system = 5
Output force from pulley system = 1,000 N
Find;
Input force of pulley system
Computation:
Mechanical advantage = Output force / Input force
Mechanical advantage of pulley system = Output force from pulley system / Input force of pulley system
5 = 1,000 / Input force of pulley system
Input force of pulley system = 1,000 / 5
Input force of pulley system = 200 N