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:
Explanation:
Resistivity is given by
where A is cross-sectional area, R is resistance, L is the length and
is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to
Answer:
Explanation:
let force exerted by engine be F.Net force =( F-400)N, applying newton law
F-400 = 1.5 x 10³x18 =27000 ,
F = 27400 N.
velocity after 12 s = 0 + 18 x 12 = 216 m/s
Average velocity = (0 + 216 )/2 = 108 m/s
Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶
b) At 12 s , velocity = 216 m/s
Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed


The force F₁is constant acceleration is also a constant.

We need to calculate the acceleration
Using formula of acceleration



Similarly,

For total force,


The speed of second tugboat is


We need to calculate total acceleration



We need to calculate the acceleration a₂



We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂



Hence, The magnitude of F₁ is 3.7 times of F₂
The correct answer is rock cycle