<span>3.2 grams
The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So
72 / 38 = 1.894736842
So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams.
The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power
= 12 g / (e ^ ln(2)*1.8947)
= 12 g / (e ^ 0.693147181 * 1.8947)
= 12 g / (e ^ 1.313305964)
= 12 g / 3.718446464
= 3.227154167 g
So rounded to 2 significant figures gives 3.2 grams.</span>
Answer:
a
![\theta = 0.0022 rad](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%200.0022%20rad)
b
![I = 0.000304 I_o](https://tex.z-dn.net/?f=I%20%20%3D%20%200.000304%20I_o)
Explanation:
From the question we are told that
The wavelength of the light is ![\lambda = 550 \ nm = 550 *10^{-9} \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20550%20%5C%20nm%20%20%3D%20%20550%20%2A10%5E%7B-9%7D%20%5C%20m)
The distance of the slit separation is ![d = 0.500 \ mm = 5.0 *10^{-4} \ m](https://tex.z-dn.net/?f=d%20%3D%200.500%20%5C%20mm%20%3D%205.0%20%2A10%5E%7B-4%7D%20%5C%20m)
Generally the condition for two slit interference is
![dsin \theta = m \lambda](https://tex.z-dn.net/?f=dsin%20%5Ctheta%20%3D%20%20m%20%5Clambda)
Where m is the order which is given from the question as m = 2
=> ![\theta = sin ^{-1} [\frac{m \lambda}{d} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7Bm%20%5Clambda%7D%7Bd%7D%20%5D)
substituting values
![\theta = 0.0022 rad](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%200.0022%20rad)
Now on the second question
The distance of separation of the slit is
![d = 0.300 \ mm = 3.0 *10^{-4} \ m](https://tex.z-dn.net/?f=d%20%3D%20%200.300%20%5C%20mm%20%20%3D%20%203.0%20%2A10%5E%7B-4%7D%20%5C%20m)
The intensity at the the angular position in part "a" is mathematically evaluated as
![I = I_o [\frac{sin \beta}{\beta} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%5Cbeta%7D%7B%5Cbeta%7D%20%5D%5E2)
Where
is mathematically evaluated as
![\beta = \frac{\pi * d * sin(\theta )}{\lambda }](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20%5Cfrac%7B%5Cpi%20%2A%20%20d%20%20%2A%20%20sin%28%5Ctheta%20%29%7D%7B%5Clambda%20%7D)
substituting values
![\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20%5Cfrac%7B3.142%20%20%2A%20%203%2A10%5E%7B-4%7D%20%20%2A%20%20sin%280.0022%20%29%7D%7B550%20%2A10%5E%7B-9%7D%20%7D)
![\beta = 0.06581](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%200.06581)
So the intensity is
![I = I_o [\frac{sin (0.06581)}{0.06581} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%280.06581%29%7D%7B0.06581%7D%20%5D%5E2)
![I = 0.000304 I_o](https://tex.z-dn.net/?f=I%20%20%3D%20%200.000304%20I_o)
Answer:
Electrical force, F = 90 N
Explanation:
It is given that,
Charge on sphere 1, ![q_1=9\ \mu C=9\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_1%3D9%5C%20%5Cmu%20C%3D9%5Ctimes%2010%5E%7B-6%7D%5C%20C)
Charge on sphere 2, ![q_1=4\ \mu C=4\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_1%3D4%5C%20%5Cmu%20C%3D4%5Ctimes%2010%5E%7B-6%7D%5C%20C)
Distance between two spheres, d = 6 cm = 0.06 m
Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,
![F=k\dfrac{q_1q_2}{d^2}](https://tex.z-dn.net/?f=F%3Dk%5Cdfrac%7Bq_1q_2%7D%7Bd%5E2%7D)
![F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}](https://tex.z-dn.net/?f=F%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B9%5Ctimes%2010%5E%7B-6%7D%5Ctimes%204%5Ctimes%2010%5E%7B-6%7D%7D%7B%280.06%29%5E2%7D)
F = 90 N
So, the electrical force between them is 90 N. Hence, this is the required solution.
Answer:
9241.6 W or 12.39318 hp
Explanation:
u = Initial velocity = 0
v = Final velocity
m = Mass
t = Time taken
Energy
![KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dm%28v%5E2-u%5E2%29%5C%5C%5CRightarrow%20KE%3D%5Cfrac%7B1%7D%7B2%7D108%2830.4%5E2-0%5E2%29%5C%5C%5CRightarrow%20KE%3D49904.64%5C%20Joules)
Power
![P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BKE%7D%7Bt%7D%5C%5C%5CRightarrow%20P%3D%5Cfrac%7B49904.64%7D%7B5.4%7D%5C%5C%5CRightarrow%20P%3D9241.6%5C%20W)
Converting to hp
![1\ W=\frac{1}{745.7}\ hp](https://tex.z-dn.net/?f=1%5C%20W%3D%5Cfrac%7B1%7D%7B745.7%7D%5C%20hp)
![\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%209241.6%5C%20W%3D%5Cfrac%7B9241.6%7D%7B745.7%7D%5C%20hp%3D12.39318%5C%20hp)
The power developed by the cheetah is 9241.6 W or 12.39318 hp