The question is incomplete. Here is the complete question.
The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.
(a) If the half-life of 238Pu is 87.7 yr, write a function of the form to model the quantity Q(t) of 238Pu left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.
(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
Answer: (a)
(b) 91 years.
Explanation:
(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:
Q(t) =
k = 0.0079
<u>Knowing k and </u><u>=3.5kg, function is </u><u />
<u />
<u>(</u>b) Using function:
t = 91.41
t ≈ 91 years
<u>Scientists will be able to receive data for approximately 91 years.</u>
1. Intrusive igneous rocks are formed from magma deep inside the Earth.
2. Intrusive rocks have large crystals that can be seen very easily. A common example of an intrusive igneous rock is granite.
3. They have large crystals because of the slow cooling process.
I tried...I hope this helps!
Answer:
Radius of the outer most dark fringe is 2.65 cm
Solution:
As per the question:
Radius of curvature of the glass, r = 10.8 m
No. of dark fringes, n = 100
Wavelength of light,
Now,
To calculate the radius R of the outermost ring:
Radius of the dark fringe of nth order is given by:
The energy of a photon when the frequency is 5 x 10¹⁴ Hz is 2.07 eV.
The energy of a photon when the frequency is 10 GHz is 4.14 x 10⁻⁵ eV.
The energy of a photon when the frequency is 30 MHz is 1.24 x 10⁻⁷ eV.
<h3>
Energy of a photon</h3>
The energy of a photon is calculated from the product of frequency and Planck's constant.
E = hf
where;
- h is Planck's constant
- f is frequency of the photon
<h3>when the frequency = 5 x 10¹⁴ Hz</h3>
E = 6.626 x 10⁻³⁴ x 5 x 10¹⁴
E = 3.313 x 10⁻¹⁹ J
E = (3.313 x 10⁻¹⁹ J) / (1.6 x 10⁻¹⁹)
E = 2.07 eV
<h3>when the frequency = 10 GHz = 10 x 10¹⁰ Hz</h3>
E = 6.626 x 10⁻³⁴ x 10 x 10¹⁰
E = 6.626 x 10⁻²⁴ J
E = (6.626 x 10⁻²⁴ J) / (1.6 x 10⁻¹⁹)
E = 4.14 x 10⁻⁵ eV
<h3>when the frequency = 30 MHz = 30 x 10⁶ Hz</h3>
E = 6.626 x 10⁻³⁴ x 30 x 10⁶
E = 1.988 x 10⁻²⁶ J
E = ( 1.988 x 10⁻²⁶ J) / (1.6 x 10⁻¹⁹)
E = 1.24 x 10⁻⁷ eV
Learn more about energy of photon here: brainly.com/question/15946945
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Answer:
= 11.8 Ω
Explanation:
Using the equation;
RT = Rr [1 + α (T -Tr)
Where;
RT = Resistance of conductor at temperature T
Rr = Resistance of conductor at reference temperature Tr
α = Temperature coefficient of resistance at reference temperature Tr
Therefore;
RT = 10 Ω [ 1 + 0.004( 65-20)]
= 10 ( 1 + 0.004(45))
= 11.8 Ω