Answer:
m =8.81*10^{-6}grams
time t = 52.8 year
Explanation:
GIVEN DATA:
the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s
activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s
activity , 
= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2
= 8.877*10^{16}
Number of moles:
n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol
mass of the CO-60 is,
m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams
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time t = -[T1/2 / ln2]*ln[N/N0]
= - [5.3 years / ln2]*ln[1x10-6/1x10-3]
= 52.8 year
Answer:
Frequency = 260 Hz and wavelength = 1.31 m
Explanation:
Given that,
A hammer begins vibrating back and forth at approximately 260 cycles per second.
(a) The frequency of an object is the number of vibrations per unit time. The frequency of the sound wave is 260 Hz.
(b) The speed of sound in air is 343 m/s. So,
Hence, this is the required solution.
Answer:
12.17 m/s²
Explanation:
The formula of period of a simple pendulum is given as,
T = 2π√(L/g)........................ Equation 1
Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie
making g the subject of the equation,
g = 4π²L/T²................... Equation 2
Given: T = 1.8 s, l = 1.00 m
Constant: π = 3.14
Substitute into equation 2
g = (4×3.14²×1)/1.8²
g = 12.17 m/s²
Hence the acceleration due to gravity of the planet = 12.17 m/s²
