The definition of density is
Density = (mass) / (volume)
Multiply each side by 'volume' : (density) x (volume) = (mass)
Divide each side by 'density' : Volume = (mass) / (density)
A. When we convert 37.4 mL to ML, the result obtained is 3.74×10¯⁸ ML
B. When we convert 689 km/hr to m/s, the result obtained is 191.39 m/s
C. When we convert 34.5 m² to mm², the result obtained is 3.45×10⁷ mm²
<h3>A. How to convert millimeters (mL) to megaliter (ML)</h3>
- Volume (mL) = 37.4 mL
- Volume (ML) =?
1 mL = 1×10¯⁹ ML
Therefore,
37.4 mL = 37.4 × 1×10¯⁹
37.4 mL = 3.74×10¯⁸ ML
Thus, 37.4 mLis equivalent to 3.74×10¯⁸ ML
<h3>B. How to convert 689 km/hr to m/s</h3>
Conversion scale
3.6 Km/hr = 1 m/s
Therefore,
689 km/hr = 689 / 3.6
689 km/hr = 191.39 m/s
Thus, 689 km/hr is equivalent to 191.39 m/s
<h3>C. How to convert 34.5 m² to mm²</h3>
Conversion scale
1 m² = 1×10⁶ mm²
Therefore,
34.5 m² = 34.5 × 1×10⁶
34.5 m² = 3.45×10⁷ mm²
Thus, 34.5 m² is equivalent to 3.45×10⁷ mm²
Learn more about conversion:
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Answer:
The sound level of the 26 geese is 
Explanation:
From the question we are told that
The sound level is 
The number of geese is 
Generally the intensity level of sound is mathematically represented as
The intensity of sound level in dB for one goose is mathematically represented as
![Z_1 = 10 log [\frac{I}{I_O} ]](https://tex.z-dn.net/?f=Z_1%20%3D%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_O%7D%20%5D)
Where I_o is the threshold level of intensity with value 
is the intensity for one goose in 
For 26 geese the intensity would be
Then the intensity of 26 geese in dB is
![Z_{26} = 10 log[\frac{26 I }{I_o} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%5B%5Cfrac%7B26%20I%20%7D%7BI_o%7D%20%5D)
![Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%2A%20%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
![Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%20%5C%20%5C%20%29%20%2A%20%20%20%28%5C%20%5C%20%2010%20log%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
From the law of logarithm we have that
![Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%2026%20%2B%20%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_0%7D%20%5D)
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
Answer:
1115560000 J
Explanation:
1/2 * 80,000 * 167^2 m/s = 1115560000 J
