Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the wheel is m = 6.9 kg
The radius is 
The radius of gyration is 
The angle is 
The force which the massless bar is subjected to 
Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.
Generally the moment of inertia about A is mathematically represented as

Here
is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

=>
=>
=>
Generally the torque experienced by the wheel is mathematically represented as

=> 
=> 
Generally this torque is also mathematically represented as

=> 
=> 
<span>The vertical axis represents the velocity of the object</span>
Answer: 272.82 drop/tile
Explanation:
Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?
Tiles/ft^2 × drop/tiles = drop/ft^2
Tiles will cancel out. Leaving the answer to be drop/ ft^2
Substitutes all the magnitude of the above units.
17 × drop/tiles = 4638
Make drop/tiles the subject of formula
Drop/tiles = 4638/17
Drop/tiles = 272.82
Therefore, 272.82 drop/tile drops fall on each tile?