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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1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:

where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:

2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c
work is distance * force so 15*100=1500
and to find time you know power = diastance * force / time
so 25=15*100/t
25=1500/t
25/1500=t
.016=time
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Cool down the Slowest
B. Seawater, because it heats up slower and gives away heat slower than sand.
Read the question wrong the first time