Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
![\frac{1}{f}=\frac{1}{u}+\frac{1}{v}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bu%7D%2B%5Cfrac%7B1%7D%7Bv%7D)
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
![\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B24%7D%2B%5Cfrac%7B1%7D%7B-71.4%7D)
![\frac{1}{f}=0.028](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D0.028)
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
![p=\frac{1}{f}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7Bf%7D)
Here, p is the power of the lens.
Put f= 35.71 cm.
![p=\frac{1}{35.71}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7B35.71%7D)
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Answer:
final temperature will be 0 degree C
Total amount of ice will be
![m_{ice} = 182 g](https://tex.z-dn.net/?f=m_%7Bice%7D%20%3D%20182%20g)
total amount of water
![m_{water} = 17 g](https://tex.z-dn.net/?f=m_%7Bwater%7D%20%3D%2017%20g)
Explanation:
After thermal equilibrium is achieved we can say that
Heat given by water = heat absorbed by ice cubes
so we will have
Heat given by water to reach 0 degree C
![Q_1 = m_1s_1 \Delta T_1](https://tex.z-dn.net/?f=Q_1%20%3D%20m_1s_1%20%5CDelta%20T_1)
![Q_1 = 0.030(4186)(50 - 0)](https://tex.z-dn.net/?f=Q_1%20%3D%200.030%284186%29%2850%20-%200%29)
![Q_1 = 6279 J](https://tex.z-dn.net/?f=Q_1%20%3D%206279%20J)
heat absorbed by ice cubes to reach 0 degree
![Q_2 = m_2 s_2 \Delta T_2](https://tex.z-dn.net/?f=Q_2%20%3D%20m_2%20s_2%20%5CDelta%20T_2%20)
![Q_2 = (0.169)(2100)(30)](https://tex.z-dn.net/?f=Q_2%20%3D%20%280.169%29%282100%29%2830%29)
![Q_2 = 10647 J](https://tex.z-dn.net/?f=Q_2%20%3D%2010647%20J)
so we will have
![Q_2 > Q_1](https://tex.z-dn.net/?f=Q_2%20%3E%20Q_1)
so here we can say that few amount of water will freeze here to balance the heat
![10647 - 6279 = mL](https://tex.z-dn.net/?f=10647%20-%206279%20%3D%20mL)
![m = \frac{10647 - 6279}{335000}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B10647%20-%206279%7D%7B335000%7D)
![m = 13 g](https://tex.z-dn.net/?f=m%20%3D%2013%20g)
so final temperature will be 0 degree C
Total amount of ice will be
![m_{ice} = 84.5 + 84.5 + 13](https://tex.z-dn.net/?f=m_%7Bice%7D%20%3D%2084.5%20%2B%2084.5%20%2B%2013)
![m_{ice} = 182 g](https://tex.z-dn.net/?f=m_%7Bice%7D%20%3D%20182%20g)
total amount of water
![m_{water} = 30 - 13](https://tex.z-dn.net/?f=m_%7Bwater%7D%20%3D%2030%20-%2013)
![m_{water} = 17 g](https://tex.z-dn.net/?f=m_%7Bwater%7D%20%3D%2017%20g)
That's wave 'diffraction'.
Answer:
this is a no brainer
Explanation:
As air pressure in an area increases, the density of the gas particles in that area increases.
(1)
Cheetah speed: ![v_c = 97.8 km/h=27.2 m/s](https://tex.z-dn.net/?f=v_c%20%3D%2097.8%20km%2Fh%3D27.2%20m%2Fs)
Its position at time t is given by
(1)
Gazelle speed: ![v_g = 78.2 km/h=21.7 m/s](https://tex.z-dn.net/?f=v_g%20%3D%2078.2%20km%2Fh%3D21.7%20m%2Fs)
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when
. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
![v_c t=S_0 + v_g t](https://tex.z-dn.net/?f=v_c%20t%3DS_0%20%2B%20v_g%20t)
![(v_c -v_g t)=S_0](https://tex.z-dn.net/?f=%28v_c%20-v_g%20t%29%3DS_0)
![t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7BS_0%7D%7Bv_c-v_t%7D%3D%5Cfrac%7B96.8%20m%7D%7B27.2%20m%2Fs-21.7%20m%2Fs%7D%3D17.6%20s)
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah: ![S_c = v_c t =(27.2 m/s)(7.5 s)=204 m](https://tex.z-dn.net/?f=S_c%20%3D%20v_c%20t%20%3D%2827.2%20m%2Fs%29%287.5%20s%29%3D204%20m)
Gazelle: ![S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m](https://tex.z-dn.net/?f=S_g%20%3D%20v_g%20t%20%3D%2821.7%20m%2Fs%29%287.5%20s%29%3D162.8%20m)
So, the gazelle should be ahead of the cheetah of at least
![d=S_c -S_g =204 m-162.8 m=41.2 m](https://tex.z-dn.net/?f=d%3DS_c%20-S_g%20%3D204%20m-162.8%20m%3D41.2%20m)