Explanation:
Types of light microscope
1. Compound , and 2. Stereo Microscope
Compound microscope has two lens system also called compound lens system. The objective lens and the eyepiece lens. The magnification provided by the objective lens is compounded by the eyepiece lens, the a higher magnification is observed.
Answer:![Q=248.011 W](https://tex.z-dn.net/?f=Q%3D248.011%20W)
Explanation:
Given
Temperature of Room ![T_{\infty }=18^{\circ}\approx 291 K](https://tex.z-dn.net/?f=T_%7B%5Cinfty%20%7D%3D18%5E%7B%5Ccirc%7D%5Capprox%20291%20K)
Area of Person ![A=1.7 m^2](https://tex.z-dn.net/?f=A%3D1.7%20m%5E2)
Temperature of skin ![T=32^{\circ}\approx 305 K](https://tex.z-dn.net/?f=T%3D32%5E%7B%5Ccirc%7D%5Capprox%20305%20K)
Heat transfer coefficient ![h=5 W/m^2.k](https://tex.z-dn.net/?f=h%3D5%20W%2Fm%5E2.k)
Emissivity of the skin and clothes ![\epsilon =0.9](https://tex.z-dn.net/?f=%5Cepsilon%20%3D0.9)
![\Delta T=32-18=14^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20T%3D32-18%3D14%5E%7B%5Ccirc%7D)
Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection
Heat transfer due radiation ![Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )](https://tex.z-dn.net/?f=Q_1%3D%5Cepsilon%20%5Csigma%20A%28T%5E4-T_%7B%5Cinfty%20%7D%5E4%20%29)
where ![\sigma =stefan-boltzman\ constant](https://tex.z-dn.net/?f=%5Csigma%20%3Dstefan-boltzman%5C%20constant)
![Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)](https://tex.z-dn.net/?f=Q_1%3D0.9%5Ctimes%205.687%5Ctimes%2010%5E%7B-8%7D%5Ctimes%201.7%5Ctimes%20%28305%5E4-291%5E4%29%20)
![Q_1=129.01 W](https://tex.z-dn.net/?f=Q_1%3D129.01%20W)
Heat Transfer due to convection is given by
![Q_2=hA(\Delta T)](https://tex.z-dn.net/?f=Q_2%3DhA%28%5CDelta%20T%29)
![Q_2=5\times 1.7\times 14=119 W](https://tex.z-dn.net/?f=Q_2%3D5%5Ctimes%201.7%5Ctimes%2014%3D119%20W)
![Q=Q_1+Q_2](https://tex.z-dn.net/?f=Q%3DQ_1%2BQ_2)
![Q=129.01+119=248.011 W](https://tex.z-dn.net/?f=Q%3D129.01%2B119%3D248.011%20W)
Answer: 2 cm
Explanation:
Given , for a converging lens
Focal length : ![f=20\ cm](https://tex.z-dn.net/?f=f%3D20%5C%20cm)
Height of object : ![h=1\ cm](https://tex.z-dn.net/?f=h%3D1%5C%20cm)
Object distabce from lens : ![u=-10\ cm](https://tex.z-dn.net/?f=u%3D-10%5C%20cm)
Using lens formula:
, we get
, where v = image distance from the lens.
On solving aboive equation , we get
![\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7B20%7D-%5Cdfrac%7B1%7D%7B10%7D%3D%5Cdfrac%7B1-2%7D%7B20%7D%3D%5Cdfrac%7B-1%7D%7B20%7D%5CRightarrow%5C%20v%3D-20%5C%20cm)
Formula of Magnification :
, where h' is the height of image.
Put value of u, v and h in it , we get
![\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm](https://tex.z-dn.net/?f=%5Cdfrac%7B-20%7D%7B-10%7D%3D%5Cdfrac%7Bh%27%7D%7B1%7D%5C%5C%5C%5C%5CRightarrow%5C%20h%27%3D2%5C%20cm)
Hence, the height of the image is 2 cm.
Answer:
c)At a distance greater than r
Explanation:
For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:
![G\frac{Mm}{r^2}=m\frac{v^2}{r}](https://tex.z-dn.net/?f=G%5Cfrac%7BMm%7D%7Br%5E2%7D%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance between the satellite and the Earth's centre
v is the speed of the satellite
Re-arranging the equation, we write
![r = \frac{GM}{v^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7BGM%7D%7Bv%5E2%7D)
so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.
Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is
c)At a distance greater than r
Answer:
it makes the object speed increase, decrease and change the direction of the object.
Hope it helps!