Answer:
JC⁻¹
Explanation:
= mass of water added to calorimeter = 94.8 g
= initial temperature of the water added = 60.4 C
= specific heat of water = 4.184 Jg⁻¹C⁻¹
= mass of water available to calorimeter = 94.8 g
= initial temperature of the water in calorimeter = 22.3 C
= final equilibrium temperature = 35 C
= Heat gained by calorimeter
Using conservation of heat
Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter
![Q_{C} = m_{a} c_{w} (T_{ai} - T_{f}) - m_{c} c_{w} (T_{f} - T_{ci})](https://tex.z-dn.net/?f=Q_%7BC%7D%20%3D%20m_%7Ba%7D%20c_%7Bw%7D%20%28T_%7Bai%7D%20-%20T_%7Bf%7D%29%20-%20%20m_%7Bc%7D%20c_%7Bw%7D%20%28T_%7Bf%7D%20-%20T_%7Bci%7D%29)
![Q_{C} = (94.8) (4.184) (60.4 - 35) - (94.8) (4.184) (35 - 22.3)](https://tex.z-dn.net/?f=Q_%7BC%7D%20%3D%20%2894.8%29%20%284.184%29%20%2860.4%20-%2035%29%20-%20%20%2894.8%29%20%284.184%29%20%2835%20-%2022.3%29)
J
= Change in temperature of calorimeter
Change in temperature of calorimeter is given as
C
Heat capacity of calorimeter is given as
![c_{cm} = \frac{Q_{C}}{T}](https://tex.z-dn.net/?f=c_%7Bcm%7D%20%3D%20%5Cfrac%7BQ_%7BC%7D%7D%7BT%7D)
![c_{cm} = \frac{5037.4}{12.7}](https://tex.z-dn.net/?f=c_%7Bcm%7D%20%3D%20%5Cfrac%7B5037.4%7D%7B12.7%7D)
JC⁻¹
Answer:
![23^{\circ}](https://tex.z-dn.net/?f=23%5E%7B%5Ccirc%7D)
Explanation:
n = Refractive index of air = 1
= Refractive index of contact lens = 1.6
= Refractive index of cornea = 1.4
= Refractive index of fluid = 1.3
From Snell's law
![n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}](https://tex.z-dn.net/?f=n%5Csin30%5E%7B%5Ccirc%7D%3Dn_1%5Csin%5Ctheta%5C%5C%5CRightarrow%20%5Ctheta%3D%5Csin%5E%7B-1%7D%5Cdfrac%7B1%5Csin30%5E%7B%5Ccirc%7D%7D%7B1.6%7D%5C%5C%5CRightarrow%20%5Ctheta%3D18.21%5E%7B%5Ccirc%7D)
![n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}](https://tex.z-dn.net/?f=n_1%5Csin%5Ctheta%3Dn_2%5Csin%5Ctheta_1%5C%5C%5CRightarrow%20%5Ctheta_%7B1%7D%3D%5Csin%5E%7B-1%7D%5Cdfrac%7B1.6%5Ctimes%20%5Csin18.21%5E%7B%5Ccirc%7D%7D%7B1.4%7D%5C%5C%5CRightarrow%20%5Ctheta_1%3D20.92%5E%7B%5Ccirc%7D)
![n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}](https://tex.z-dn.net/?f=n_2%5Csin%5Ctheta_1%3Dn_3%5Csin%5Ctheta_3%5C%5C%5CRightarrow%20%5Ctheta_3%3D%5Csin%5E%7B-1%7D%5Cdfrac%7B1.4%5Csin20.92%5E%7B%5Ccirc%7D%7D%7B1.3%7D%5C%5C%5CRightarrow%20%5Ctheta_3%3D22.62%5E%7B%5Ccirc%7D%5Capprox%2023%5E%7B%5Ccirc%7D)
The angle is the light traveling in the fluid behind her cornea is
.
Answer:
whats the question...............
Explanation:
Here's my guesses.
1 sea stack ... A2 inlet ... E3 sand spit ... F4 sea arch ... B (looks like a land arch made by the sea ... ?)5 lagoon ... C6 barrier island ...A7 bay ... D8 sea cliff .... J9 sea cave .... G10 tombolo .... no idea
That's a lot of work for 9 points !!!!
Did you forget to add the picture or is this a free point