Answer:
49.3 N
Explanation:
Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?
The weight of the bucket of water = mg.
Weight = 4.25 × 9.8
Weight = 41.65 N
The tension and the weight will be opposite in direction.
Total force = ma
T - mg = ma
Make tension T the subject of formula
T = ma + mg
T = m ( a + g )
Substitutes all the parameters into the formula
T = 4.25 ( 1.8 + 9.8 )
T = 4.25 ( 11.6 )
T = 49.3 N
Therefore, the tension in the rope is 49.3 N approximately.
The lateral displacement is I don’t know tbh I think 16.8
<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:
15.2 s
Explanation:
Convert hp to W:
55.0 hp × 746 W/hp = 41,030 W
Power = energy / time
41030 W = 6.22×10⁵ J / t
t = 15.2 s
- Initial velocity (u) = 0 m/s [the car was at rest]
- Distance (s) = 80 m
- Time (t) = 10 s
- Let the magnitude of acceleration be a.
- By using the equation of motion,
we get,
<u>A</u><u>nswer:</u>
<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.