The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg? SOLUTION (a) Let m(t) be the mass of radium-226 (in milligrams) that remains after t years. Then dm/dt = km and m(0) = 100, so this theorem gives m(t) = m(0)ekt = ekt. In order to determine the value of k, we use the fact that m(1590) = 1 2 . Thus e1590k = so e1590k = and 1590k = ln 1 2 = − ln(2) k = . Therefore m(t) = . We could use the fact that eln(2) = 2 to write the expression for m(t) in the alternative form m(t) = . (b) The mass after 1000 years is as follows. (Round your final answer to the nearest milligram.) m(1000) = ≈ mg

See explanation

Explanation:

a) Formula for the mass of the sample that remains after t years = N= Noe^-kt

Where;

N = mass at time t years

No = mass at time t= 0

k = decay constant

t = time taken

So,

N = 100e^-kt

b) First,

t1/2 = -ln(1/2)/k

t1/2 = 0.693/k

t1/2 = half life of radium-226 =1590 years

1590 = 0.693/k

k = 0.693/1590

k = 4.36 * 10^-4

So,

N= 100e^-(4.36 * 10^-4 * 1000)

N= 65 mg

c) From

N = 100e^-kt

40 = 100e^-(4.36 * 10^-4t)

40/100 = e^-(4.36 * 10^-4t)

0.4 = e^-(4.36 * 10^-4t)

ln(0.4) = ln(e^-(4.36 * 10^-4t))

-0.9163 = -4.36 * 10^-4t

t = 0.9163/4.36 * 10^-4

t = 2102 years

prolly a beacuse the momentuem the car has increases while going down hill

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.