Exactly the same
Explanation:
Objects looks exactly the same when viewed through a transparent object. A transparent body allows light to pass through completely. Therefore, the image formed will completely resemble that of the object to be viewed.
Examples are air, glass, pure water.
A translucent body partially allows light to pass through it. Most times, the image is not clear and it is distorted.
An opaque object will not allow light pass through it. It completely obstructs the ray of path of light. An opaque body will produce an image on the screen.
Learn more:
Refraction brainly.com/question/12370040
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Answer:
A
The current through the solution is ![I =35.61mA](https://tex.z-dn.net/?f=I%20%3D35.61mA)
B
The current is moving from B to A
Explanation:
From the question we are told that
The number of
that move from A to B is ![N_{cl} = 5.02 *10^{18} ions](https://tex.z-dn.net/?f=N_%7Bcl%7D%20%3D%205.02%20%2A10%5E%7B18%7D%20ions)
The time taken to move from A to B ![t_{cl} = 0.620](https://tex.z-dn.net/?f=t_%7Bcl%7D%20%3D%200.620)
![=37.2 s](https://tex.z-dn.net/?f=%3D37.2%20s)
Since the value of 1 charge is ![q =1.602 *10^{-19} C](https://tex.z-dn.net/?f=q%20%3D1.602%20%2A10%5E%7B-19%7D%20C)
The quantity of charge Q that flow from A to B is mathematically given as
![Q_{cl} = 5.02 *10^{18} * 1.602*10^{-19}](https://tex.z-dn.net/?f=Q_%7Bcl%7D%20%3D%205.02%20%2A10%5E%7B18%7D%20%2A%201.602%2A10%5E%7B-19%7D)
The number of
that move from A to B is ![N_{ca} = 3.25*10^{18} ions](https://tex.z-dn.net/?f=N_%7Bca%7D%20%3D%203.25%2A10%5E%7B18%7D%20ions)
Since time taken to move from A to B is equal to time taken to move from B to A ![t =t_{cl} =t_{ca}= 0.620=37.2s](https://tex.z-dn.net/?f=t%20%3Dt_%7Bcl%7D%20%3Dt_%7Bca%7D%3D%200.620%3D37.2s)
The quantity of charge Q that flow from B to A is mathematically given as
![Q_{ca} = 3.25 *10^{18} * 1.602*10^{-19}](https://tex.z-dn.net/?f=Q_%7Bca%7D%20%3D%203.25%20%2A10%5E%7B18%7D%20%2A%201.602%2A10%5E%7B-19%7D)
![=0.5207C](https://tex.z-dn.net/?f=%3D0.5207C)
The total quantity of charge is
![Q_{tot}=Q_{cl} + Q_{ca}](https://tex.z-dn.net/?f=Q_%7Btot%7D%3DQ_%7Bcl%7D%20%2B%20Q_%7Bca%7D)
![Q_{tot}= 0.804 + 0.5207](https://tex.z-dn.net/?f=Q_%7Btot%7D%3D%200.804%20%2B%200.5207)
![= 1.325C](https://tex.z-dn.net/?f=%3D%201.325C)
The current flowing through the solution is
![I =\frac{Q_{tot}}{t}](https://tex.z-dn.net/?f=I%20%3D%5Cfrac%7BQ_%7Btot%7D%7D%7Bt%7D)
![I = \frac{1.325}{37.2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1.325%7D%7B37.2%7D)
![= 0.03561A](https://tex.z-dn.net/?f=%3D%200.03561A)
![I =35.61mA](https://tex.z-dn.net/?f=I%20%3D35.61mA)
The flow is from B to A cause current flow from the positive terminal to negative terminal
Answer:
![MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093](https://tex.z-dn.net/?f=%20MA_1%20%3D%20%5Cfrac%7B50%20m%2Fs%7D%7B%5Csqtr%7B1.288%2A188.9%20J%2FKg%20K%20%2A%201200%20K%7D%7D%3D0.093)
![MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73](https://tex.z-dn.net/?f=%20MA_2%20%3D%5Cfrac%7B1163.074%20m%2Fs%7D%7B%5Csqrt%7B1.288%20%2A188.9%20J%2FKg%20K%20%2A%20400%20K%7D%7D%3D3.73)
Explanation:
Assuming this problem: "Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat assumption."
Part a
For this case we can assume at the inlet we have the following properties:
![T_1 = 1200 K, v_1 = 50 m/s](https://tex.z-dn.net/?f=%20T_1%20%3D%201200%20K%2C%20v_1%20%3D%2050%20m%2Fs%20)
We can calculate the Mach number with the following formula:
![MA_1 = \frac{v_1}{c_1} = \frac{v_1}{\sqrt{kRT}}](https://tex.z-dn.net/?f=%20MA_1%20%3D%20%5Cfrac%7Bv_1%7D%7Bc_1%7D%20%3D%20%5Cfrac%7Bv_1%7D%7B%5Csqrt%7BkRT%7D%7D)
Where k represent the specific ratio given k =1.288 and R would be the universal gas constant for the carbon diaxide given by: ![R= 188.9 J/ Kg K](https://tex.z-dn.net/?f=%20R%3D%20188.9%20J%2F%20Kg%20K)
And if we replace we got:
![MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093](https://tex.z-dn.net/?f=%20MA_1%20%3D%20%5Cfrac%7B50%20m%2Fs%7D%7B%5Csqtr%7B1.288%2A188.9%20J%2FKg%20K%20%2A%201200%20K%7D%7D%3D0.093)
Part b
For this case we can use the same formula:
![MA_2 = \frac{v_2}{c_2}](https://tex.z-dn.net/?f=%20MA_2%20%3D%20%5Cfrac%7Bv_2%7D%7Bc_2%7D%20)
And we can obtain the value of v2 from the total energy of adiabatic flow process, given by this equation:
![c_p T_1 + \frac{v^2_1}{2}=c_p T_2 + \frac{v^2_2}{2}](https://tex.z-dn.net/?f=%20c_p%20T_1%20%2B%20%5Cfrac%7Bv%5E2_1%7D%7B2%7D%3Dc_p%20T_2%20%2B%20%5Cfrac%7Bv%5E2_2%7D%7B2%7D)
The value of
and the value fo T2 = 400 K so we can solve for
and we got:
![v_2= \sqrt{2c_p (T_1 -T_2) +v^2_1}=1163.074 m/s](https://tex.z-dn.net/?f=%20v_2%3D%20%5Csqrt%7B2c_p%20%28T_1%20-T_2%29%20%2Bv%5E2_1%7D%3D1163.074%20m%2Fs)
And now we can replace on this equation:
![MA_2 = \frac{v_2}{c_2}](https://tex.z-dn.net/?f=%20MA_2%20%3D%20%5Cfrac%7Bv_2%7D%7Bc_2%7D%20)
And we got:
![MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73](https://tex.z-dn.net/?f=%20MA_2%20%3D%5Cfrac%7B1163.074%20m%2Fs%7D%7B%5Csqrt%7B1.288%20%2A188.9%20J%2FKg%20K%20%2A%20400%20K%7D%7D%3D3.73)
Answer:
It will have an acceleration of
.
Explanation:
Newton’s second law is defined as:
(1)
Equation 1 can be rewritten for the case of the acceleration:
The force and the acceleration are directly proportional (if one increase the other will increase too).
![a = \frac{5.0 N}{7.0 Kg}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B5.0%20N%7D%7B7.0%20Kg%7D)
But 1 N is equivalent to ![1 Kg.m/s^{2}](https://tex.z-dn.net/?f=1%20Kg.m%2Fs%5E%7B2%7D)
![a = \frac{5.0 Kg.m/s^{2}}{7.0 Kg}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B5.0%20Kg.m%2Fs%5E%7B2%7D%7D%7B7.0%20Kg%7D)
So a body will have an acceleration as a consequence of the action of a force.
Answer:
a. Find the numerical value of the change of the electric potential energy
Explanation: