Answer:
The focal length of the lens is 5.54 cm and the height of the image is -0.89 cm.
Explanation:
Given that,
Height of object h = 2 cm
Object distance u= -18 cm
Image distance v= 8 cm
We need to calculate the focal length of the lens
Using formula of lens
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D)
Where, f = focal length
Put the value into the formula
![\dfrac{1}{f}=\dfrac{1}{8}-\dfrac{1}{-18}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7B8%7D-%5Cdfrac%7B1%7D%7B-18%7D)
![\dfrac{1}{f}=\dfrac{13}{72}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B13%7D%7B72%7D)
![f=5.54\ cm](https://tex.z-dn.net/?f=f%3D5.54%5C%20cm)
(II). We need to calculate the height of the image
Using formula of magnification
![m =\dfrac{v}{u}=\dfrac{h'}{h}](https://tex.z-dn.net/?f=m%20%3D%5Cdfrac%7Bv%7D%7Bu%7D%3D%5Cdfrac%7Bh%27%7D%7Bh%7D)
![\dfrac{v}{u}=\dfrac{h'}{h}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%7D%7Bu%7D%3D%5Cdfrac%7Bh%27%7D%7Bh%7D)
Put the value into the formula
![\dfrac{8}{-18}=\dfrac{h'}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B8%7D%7B-18%7D%3D%5Cdfrac%7Bh%27%7D%7B2%7D)
![h'=\dfrac{8}{-18}\times2](https://tex.z-dn.net/?f=h%27%3D%5Cdfrac%7B8%7D%7B-18%7D%5Ctimes2)
![h'=-0.89\ cm](https://tex.z-dn.net/?f=h%27%3D-0.89%5C%20cm)
Hence, The focal length of the lens is 5.54 cm and the height of the image is -0.89 cm.
Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
For a gas
![P_{1}V_{1}=nRT_{1}](https://tex.z-dn.net/?f=P_%7B1%7DV_%7B1%7D%3DnRT_%7B1%7D)
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)
![\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7B1%7DV%7D%7BP_%7B2%7DV%7D%3D%5Cdfrac%7BnRT_%7B1%7D%7D%7B%5Cdfrac%7Bn%7D%7B2%7DRT_%7B2%7D%7D)
![\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}](https://tex.z-dn.net/?f=%5Cdfrac%7B10%5Ctimes%20V%7D%7BP_%7B2%7DV%7D%3D%5Cdfrac%7BnR%5Ctimes286%7D%7B%5Cdfrac%7Bn%7D%7B2%7DR368%7D)
![P_{2}=\dfrac{10\times368}{2\times286}](https://tex.z-dn.net/?f=P_%7B2%7D%3D%5Cdfrac%7B10%5Ctimes368%7D%7B2%5Ctimes286%7D)
![P_{2}= 6.433\ atm](https://tex.z-dn.net/?f=P_%7B2%7D%3D%206.433%5C%20atm)
![P_{2}=6.4\ atm](https://tex.z-dn.net/?f=P_%7B2%7D%3D6.4%5C%20atm)
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
<h2>
a) Volume of anchor is 0.019 m³</h2><h2>
b) Weight of anchor in air is 1463.17 N</h2>
Explanation:
a) Weight loss in water = Volume of object x density of water x acceleration due to gravity
183 = Volume of anchor x 1000 x 9.81
Volume of anchor = 0.019 m³
b) Weight = Mass x acceleration due to gravity = Volume x Density x acceleration due to gravity
Weight of anchor in air = Volume of anchor x Density of anchor x acceleration due to gravity
Weight of anchor in air = 0.019 x 7850 x 9.81 = 1463.17 N
Weight of anchor in air = 1463.17 N