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:
A. Velocity
Explanation:
Velocity is vector quantity thus has both magnitude and direction. It describes not only the speed but also the direction. Speed is scalar quantity so describes only speed but not direction. Energy has nothing to do with speed, acceleration describes change in velocity in a direction over time
Answer:
0.79 s
Explanation:
We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that 
Now, we find the acceleration:

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

Answer:
A = 4.6 [m²]
Explanation:
The area of a circle can be calculated by means of the following equation.

where:
A = area [m²]
D = diameter = 2.42 [m]
Now replacing:
![A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%2A%282.42%29%5E%7B2%7D%20%5C%5CA%20%3D%204.6%20%5Bm%5E%7B2%7D%20%5D)
Answer:
16∠45° Ω
Explanation:
Applying,
Z = V/I................... Equation 1
Where Z = Impedance, V = Voltage output, I = current input.
Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°
Substitute these values into equation 1
Z = 120cos(10t+75°)/7.5cos(10t+30)
Z = 120∠75°/ 7.5∠30°
Z = 16∠(75°-30)
Z = 16∠45° Ω
Hence the impedance of the linear network is 16∠45° Ω