The string that will break first depends on the weight of the block or bar subjected to such string.
According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.
F= kx
where;
- <em>F is the force applied to the object = weight of the object</em>
For given two identical strings, the string that will break first depends on the mass of the bar and the block.
- If the bar is heavier than the block, then the string subjected to the bar will break first.
- On the other hand, if the block is heavier than the bar, then the string subjected to the block will break first.
Thus, the string that will break first depends on the weight of the block or bar subjected to such string.
Learn more here:brainly.com/question/4404276
A) that's not physics, that's chemistry B) assuming you want it balanced, 1,2,1,2
1.41 × 10³⁰ MeV
As we know, E=mc², where E is energy, m is mass and c is the speed of light(i.e. 3×10⁸ m/s).
Given mass = 2.5 kgs
∴ E = (2.5)×(3×10⁸)² J = 22.5×10¹⁶ J
As our answer is in joules so we have to convert it into mega electron volt(MeV)
1 J = 6.242 × 10¹² MeV
⇒ 22.5×10¹⁶ J = 22.5×10¹⁶ × (6.242 × 10¹²) MeV
⇒1.41 × 10³⁰ MeV
If you want to learn more about mass-energy conversions then you can check out this link:
https://brainly.in/question/9760064
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
