<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
Answer:
option C
Explanation:
given,
force act on west = 20 lb
force act at 45° east of north = 80 lb
magnitude of force = ?
∑ F y = 80 cos 45⁰
F y = 56.57 lb
magnitude of forces in x- direction
∑ F x = -20 + 80 sin 45⁰
= 36.57 lb
net force
F = 
F = 
F = 67.36 lb≅ 67 lb
hence, the correct answer is option C
That would be a maximum of 4 atoms
The radius of the prop blade of an airplane is determined as 4.25 m.
<h3>
Radius of the prop blade</h3>
The radius of the prop blade of an airplane is calculated as follows;
a = v²/r
where;
- v is the linear speed
- r is the radius of the prop blade
- a is the centripetal acceleration
r = v²/a
r = (875²)/(180,000)
r = 4.25 m
Thus, the radius of the prop blade of an airplane is determined as 4.25 m.
Learn more about centripetal acceleration here: brainly.com/question/79801
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