Below are the choices that can be found from other source:
a. 0.3 AU
<span>b. 3.8 AU </span>
<span>c. 30 AU </span>
<span>d. 3 million AU
</span>
Answer is c. 30 AU.
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Answer: Option (c) is the correct answer.
Explanation:
It is known that the relation between resistance, length and cross-sectional area is as follows.
R = ![\rho \frac{l}{A}](https://tex.z-dn.net/?f=%5Crho%20%5Cfrac%7Bl%7D%7BA%7D)
Let the resistance of resistor A is denoted by R and the resistance of resistor B is denoted by R'.
Hence, for resistor A the expression for resistance according to the given data is as follows.
R = ![\rho \frac{2l}{2A}](https://tex.z-dn.net/?f=%5Crho%20%5Cfrac%7B2l%7D%7B2A%7D)
On cancelling the common terms we get the expression as follows.
R = ![\rho \frac{l}{A}](https://tex.z-dn.net/?f=%5Crho%20%5Cfrac%7Bl%7D%7BA%7D)
Now, the resistance for resistor B is as follows.
R' = ![\rho \frac{l'}{A'}](https://tex.z-dn.net/?f=%5Crho%20%5Cfrac%7Bl%27%7D%7BA%27%7D)
Thus, we can conclude that the statement, Wire A has the same resistance as wire B, accurately compares the resistances of wire resistors A and B.
Answer:
The answer is "
"
Explanation:
Calculating the mass flow rate of fluid:
![m= \rho AV](https://tex.z-dn.net/?f=m%3D%20%5Crho%20AV)
![= \rho \frac{\pi}{4} D^2\ V\\\\= 100 \times \frac{\pi}{4} \times (0.0127)^2\times 0.2\\\\=0.0253 \ \frac{kg}{s}\\\\](https://tex.z-dn.net/?f=%3D%20%5Crho%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20D%5E2%5C%20V%5C%5C%5C%5C%3D%20100%20%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20%280.0127%29%5E2%5Ctimes%200.2%5C%5C%5C%5C%3D0.0253%20%5C%20%5Cfrac%7Bkg%7D%7Bs%7D%5C%5C%5C%5C)
Calculating the amount of heat transfer.
![q =m\timesC_p(T_{m,0}-T_{m,i})](https://tex.z-dn.net/?f=q%20%3Dm%5CtimesC_p%28T_%7Bm%2C0%7D-T_%7Bm%2Ci%7D%29)
![=0.0253 \times 4000 (75-25)\\\\=0.0253 \times 4000(50)\\\\=0.0253 \times 200,000 \\\\= 5060 \ W](https://tex.z-dn.net/?f=%3D0.0253%20%5Ctimes%204000%20%2875-25%29%5C%5C%5C%5C%3D0.0253%20%5Ctimes%204000%2850%29%5C%5C%5C%5C%3D0.0253%20%5Ctimes%20200%2C000%20%5C%5C%5C%5C%3D%205060%20%5C%20W)
Calculating the required value for heat flux:
![q"=\frac{q}{A_s}](https://tex.z-dn.net/?f=q%22%3D%5Cfrac%7Bq%7D%7BA_s%7D)
![=\frac{q}{\pi DL}\\\\= \frac{5060}{\pi \times 0.0127 \times 10}\\\\= 12682.267 \frac{W}{m^2}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bq%7D%7B%5Cpi%20DL%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B5060%7D%7B%5Cpi%20%5Ctimes%200.0127%20%5Ctimes%2010%7D%5C%5C%5C%5C%3D%2012682.267%20%5Cfrac%7BW%7D%7Bm%5E2%7D%5C%5C)
Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
![x_{1}=\dfrac{\lambda D}{2d}](https://tex.z-dn.net/?f=x_%7B1%7D%3D%5Cdfrac%7B%5Clambda%20D%7D%7B2d%7D)
Second minima from center
![x_{2}=\dfrac{3\lambda D}{2d}](https://tex.z-dn.net/?f=x_%7B2%7D%3D%5Cdfrac%7B3%5Clambda%20D%7D%7B2d%7D)
The distance between the places where the intensity is zero due to the double slit effect
![\Delta x_{d}=x_{2}-x_{1}](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3Dx_%7B2%7D-x_%7B1%7D)
![\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3D%5Cdfrac%7B3%5Clambda%20D%7D%7B2d%7D-%5Cdfrac%7B%5Clambda%20D%7D%7B2d%7D)
![\Delta x_{d}=\dfrac{\lambda D}{d}](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3D%5Cdfrac%7B%5Clambda%20D%7D%7Bd%7D)
Put the value into the formula
![\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3D%5Cdfrac%7B600%5Ctimes10%5E%7B-9%7D%5Ctimes1%7D%7B0.04%5Ctimes10%5E%7B-3%7D%7D)
![\Delta x_{d}=0.015 =15\times10^{-3}\ m](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3D0.015%20%3D15%5Ctimes10%5E%7B-3%7D%5C%20m)
![\Delta x_{d}=15\ mm](https://tex.z-dn.net/?f=%5CDelta%20x_%7Bd%7D%3D15%5C%20mm)
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
Explanation:
If F is the applied force and θ is the applied angle from horizontal with positive angle below the horizon
Fnet = Fcosθ - μ(mg + Fsinθ)
The only variable you've actually given us is the friction coefficient.
You'll have to plug your own numbers in.
Remember, if the applied force is acting towards the floor, the normal force is increased as will be the friction force. The net force will decrease.
If the applied force is acting upward, the θ angle will be negative and the normal force is decreased along with the friction force. The net force will increase.