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:
453 gm
Explanation:
<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
Resultant is 5 m/s using the Pythagorean theorem<span />
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
C I think lol but I have to keep typing
