The top row of boxes is " F O R C E " .
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as



A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
<h3>8. </h3>
m= 0. 113 kg
W = 1.1074 N
g = 9.8 m/ s²
<h3>9. </h3>
m= 870 kg
W = 8526 N
g = 9.8 m/s²
<h3>10.</h3>
a) weight on earth:
weight = mass × acceleration due to gravity (g)
<em>g</em><em> </em><em>on</em><em> </em><em>earth</em><em> </em><em>=</em><em> </em><em>9</em><em>.</em><em>8</em><em> </em><em>m</em><em>/</em><em>s</em><em>²</em>
W = 75 × 9.8
<u>W</u><u> </u><u>= </u><u>7</u><u>3</u><u>5</u><u> </u><u>N</u>
b) g on Mars
285 = 75 × g
g = 285/ 75
<u>g = 3.8 m/ s²</u>
With the Pythagoras Theorem we can find that the distance traveled is:
39 m
The distance is the length of the path between two points and it is a scalar magnitude so if the path changes direction the Pythagorean theorem should be used
d =
Where d is the distance, (x,y,z) is the interes point, (x₀,y₀,z₀) is de reference point.
In this case, let's set a reference system in the lower part of the school, take the z-axis as vertical and set the point of arrival at as the reference (0, 0, 0).
The distance that the students descend is d₁ = 30 k ^ m, when they arrive from the bottom of the school they travel d₂ = 15 j ^ m and d₃ = 20 i ^ m
let's calculate
d =
d = 39.05 m
Notice that the distance by being a scalar does not have unit vectors
In conclusion using the Pythagoras Theorem we can find that the distance traveled is 39 m
Learn more about distance here:
brainly.com/question/7942332