Answer:
50 m/s opposite direction to the motion of the truck
Explanation:
From the question,
Applying the law of conservation of momentum
mu+m'u' = V(m+m')...….. Equation 1
Where m = mass of the truck, u = initial velocity of the truck, m' = mass of the car, u' = initial velocity of the car, V = Final velocity after collision
Given: m = 2500 kg, u = 20 m/s, m' = 1000 kg, V = 0 m/s (both car stop after collision)
Substitute these values into equation 1
2500(20)+1000(u') = 0(2500+1000)
2500(20)+1000(u') = 0
Solve for u'
u' = -[2500(20)]/1000
u' = -50 m/s
The negative sign shows that the car travels in opposite direction to the truck
Hence the car initial velocity before collision is 50 m/s in opposite direction to the motion of the truck
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:
Poynting vector, ![S=1.03\times 10^{7}\ W/m^2](https://tex.z-dn.net/?f=S%3D1.03%5Ctimes%2010%5E%7B7%7D%5C%20W%2Fm%5E2)
Explanation:
It is given that, The electric and magnetic fields of an electromagnetic wave are given by :
Electric field, ![E=6.25\times 10^{-3}\ V/m](https://tex.z-dn.net/?f=E%3D6.25%5Ctimes%2010%5E%7B-3%7D%5C%20V%2Fm)
Magnetic field, ![B=2.08\times 10^{-11}\ T](https://tex.z-dn.net/?f=B%3D2.08%5Ctimes%2010%5E%7B-11%7D%5C%20T)
We need to find the Poynting vector for this wave. the Poynting vector is given by :
![S=\dfrac{1}{\mu_o}EB](https://tex.z-dn.net/?f=S%3D%5Cdfrac%7B1%7D%7B%5Cmu_o%7DEB)
![S=\dfrac{1}{4\pi \times 10^{-7}}\times 6.25\times 10^{-3}\times 2.08\times 10^{-11}](https://tex.z-dn.net/?f=S%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D%5Ctimes%206.25%5Ctimes%2010%5E%7B-3%7D%5Ctimes%202.08%5Ctimes%2010%5E%7B-11%7D)
![S=1.03\times 10^{7}\ W/m^2](https://tex.z-dn.net/?f=S%3D1.03%5Ctimes%2010%5E%7B7%7D%5C%20W%2Fm%5E2)
So, the Poynting vector for this wave is
. Hence, this is the required solution.
Answer:
the force would increase 4 times more
Explanation
more force results more mass or acceleration