<span>A pulse with an amplitude of 3+ would be considered as increased.
Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale.
0 = absent
1+ = Weak/thready pulse
2+ Normal Pulse
3+ = Full, firm pulse.
from the above scale we can find that the 3+ reading shows that the pulse is increased.</span>
Answer:
490 in^3 = 8.03 L
Explanation:
Given:
The engine displacement = 490 in^3
= 490 in³
To determine the engine piston displacement in liters L;
(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)
First, we will convert in³ to cm³
Since 1 in = 2.54 cm
∴ 1 in³ = 16.387 cm³
If 1 in³ = 16.387 cm³
Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³
∴ 490 in³ = 8029.63 cm³
Now will convert cm³ to dm³
(NOTE: 1 L = 1 dm³)
1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm
∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³
If 1 cm³ = 1 × 10⁻³ dm³
Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³
≅ 8.03 dm³
∴ 8029.63 cm³ = 8.03 dm³
Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³
Since 1L = 1 dm³
∴ 8.03 dm³ = 8.03 L
Hence, 490 in³ = 8.03 L
<u>Answer:</u>
The correct answer option is C. 2.
<u>Explanation:</u>
We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.
According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.
Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.
It is called exothermic reaction because it releases heat and light and it is called combustion reaction because it is reacting and is being oxidised by O2 to MgO.
It can also be called as oxidation reaction since Mg is oxidised to MgO.
The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:
Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.
