Answer:
Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.
Listen if you have to cheat for the dba thing not worth even doing that dba tbh this is very easy as I just did it in like 5 minutes it gives you everything you need even the formulas so use your f .u .c .k. 1 .n g brain you monkey...
Answer:
L= 12 light years
Explanation:
for length dilation we use the formula
now calculating Lo
Lo = 12.5×365×24×3600×3×10^8
= 1.183×10^17 m
now putting the values of v and Lo in the above equation we get
= 1.136×10^17 m
L= 
m
so L= 12 light years
Explanation:
Formula for calculating the area of a rectangle A = Length *width
For statement A;
Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
Area of the rectangle = 2.536mm * 1.4mm
Area of the rectangle = 3.5504mm²
The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.
Area of the rectangle = 3.6mm² (to 2sf)
For statement B;
Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.
Area of the rectangle = 2.536mm * 1.41mm
Area of the rectangle = 3.57576mm²
Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.
Area of the rectangle = 3.58mm² (to 3sf)
Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.
Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 ×
m
diameter d = 0.074 cm = 74 ×
m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W
