Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
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Answer:
try to pop it back in good luck im scared for you
Answer:
Option A (energy transformations) is the correct solution.
Explanation:
- Energy has become a local matter that has been primarily described as the commitment to deal as well as to generate heat. There are several other power sources, which can either be transferred or transformed to several other renewable technologies.
- This same fundamental law defined that the energy required including its universe has been corrected and therefore that energy is being generated as well as severely damaged but should only be moved from one location to the next.
Those certain possibilities are not connected to the circumstance in question. And therefore this answer seems to be the correct one.
Answer:
The maximum length of the specimen before the deformation was 358 mm or 0.358 m.
Explanation:
The specific deformation ε for the material is:
(1)
Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:
(2)
Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:
So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:
Then de specific defotmation:
Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:
