Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
The longest wavelength in the Molecule's absorption spectrum is 2250nm
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Answer: you want your input force harder
Explanation:
Here we have to add the two measurements given in the question
The measurement values are given as 1.0090 cm and 0.02 cm.we have to add them on the basis of significant figure rules.
As per the addition rule in terms of significant figures
1-First we have to select the number of significant digits after the decimal point of each quantity.
2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.
3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division
Now we have to add these two quantities as per the above rule-
1.0090 cm +0.02 cm
=1.0290 cm
Here the result will follow 0.02 which has minimum number of significant figures after the decimal points.
Hence we have to round off the number from 9 of 1.0290
As 9 is greater than 5 ,so he actual result will be 1.03 cm
