Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
The fraction of radioisotope left after 1 day is
, with the half-life expressed in days
Explanation:
The question is incomplete: however, we can still answer as follows.
The mass of a radioactive sample after a time t is given by the equation:

where:
is the mass of the radioactive sample at t = 0
is the half-life of the sample
This means that the mass of the sample halves after one half-life.
We can rewrite the equation as

And the term on the left represents the fraction of the radioisotope left after a certain time t.
Therefore, after t = 1 days, the fraction of radioisotope left in the body is

where the half-life
must be expressed in days in order to match the units.
Learn more about radioactive decay:
brainly.com/question/4207569
brainly.com/question/1695370
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The properti that you didnt share
Answer:
E= -3.166 cosωt V
Explanation:
Given that
I = Imax sinωt
L= 8.4 m H
Imax= 4 A
f = ω/2π = 60.0 Hz
ω = 120π rad/s
We know that self induce E given as




E= -3166.72 cosωt m V
E= -3.166 cosωt V
This is the induce emf.