After one day, the rate of increase in Delta Cephei's brightness is;0.46
We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;
B(t)=4.0+3.5 sin(2πt/5.4)
Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides
B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)
Currently, at t = 1, we have;
B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)
Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:
B'(1)=(2π/5.4)×0.3961
B'(1)≈0.46
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Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
Answer:
1280 mg
Explanation:
Radioactive decay is a phenomenon that occurs when a certain isotope of an element, said to be radioactive, decays, turning into a lighter nucleus and emitting radiation + energy in the process.
The radioactive decay of this isotope of polonium is described by the equation
where
is the amount of polonium left after time t
is the amount of polonium t time t = 0
is the half-life of the polonium, in days (it is the time it takes for the initial element to halve its amount)
IN this problem, we know that:
After t = 560 days, the amount of polonium left is 
. Therefore, we can re-arrange the equation, substituting t = 560 d, and solve for to find the initial amount of polonium:
