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
If you were given distance & period of time, you would be able to calculate the speed.
Hope this helps!
The wavelength is 2m.
Hence, Option c) 2m is the correct answer
Given that;
Frequency; 
Speed; 
Wavelength; 
using the expression for the relations between wavelength, frequency and speed of wave:
Where 
is wavelength, f is frequency and v is speed.
We substitute our given values into the equation
The wavelength is 2m.
Hence, Option c) 2m is the correct answer.
To learn more about wavelength, click here: brainly.com/question/1347107
Answer:
40m
Explanation:
let's calculate the acceleration first
force = mass × acceleration
rearranging to find acceleration:
acceleration = force ÷ mass
force = 25N, mass = 5.0kg
acceleration = 25 ÷ 5 = 5ms^-2
we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance
rearranging v^2 = u^2 + 2as the distance is
s = (v^2 - u^2) ÷ 2a
v = 20, u = 0, a = 5
s = (20^2 - 0^2) ÷ (2 × 5) = 40m
the distance is 40m
