A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
Answer:
C
Explanation:
If Ami is saying she likes it then it it personal. If you are speaking from statistics and studies it is impersonal and technically not from there perspective. All of these do this except C.
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is 
Clock time of computer B is 
Effective CPI of computer A is 
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by 
Computer A is 1.41 times faster than the Computer B
Only 10 Kg of secondary consumers can be supported by the ecosystem.
There exists such a thing as the 10% rule in the transfer of energy in biomass from one trophic level to another. This rule ultimately controls the mass of biomass that is available at the next higher trophic level.
This rule state that only 10% of the energy in biomass is transferred to the next trophic level. By this rule, only 10% of the energy in the plant material is transferred to the carnivore population. Therefore, only 10 Kg of secondary consumers can be supported by the ecosystem.
Learn more about trophic level: brainly.com/question/13267087
