Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm
The answer is a.12.5kg because i just did the test and it was correct.
hope this helps
Answer:
(a) r = 1.062·R
= 
(b) r = 
(c) Zero
Explanation:
Here we have escape velocity v
given by
and the maximum height given by

Therefore, when the initial speed is 0.241v
we have
v =
so that;
v² =
v² = 
is then

Which gives
or
r = 1.062·R
(b) Here we have

Therefore we put
in the maximum height equation to get

From which we get
r = 1.32·R
(c) The we have the least initial mechanical energy, ME given by
ME = KE - PE
Where the KE = PE required to leave the earth we have
ME = KE - KE = 0
The least initial mechanical energy to leave the earth is zero.
The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:

where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get


Therefore. the elastic potential energy stored in the spring is 6.75J .
Learn more about potential energy here:
brainly.com/question/10770261
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There are six steps to this process , I uploaded step one and as you can see you can get all six on Quizlet:). Good luck