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Scilla [17]
4 years ago
15

Is a liter the same as a cubic meter

Physics
1 answer:
Anna71 [15]4 years ago
6 0
No.  They're both units of volume, but
the cubic meter has 1,000 liters in it.
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Convert mm3 into m3.​
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1 \times 10 { - }^{9}

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What effect does the sun have on the earth. written response
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4 years ago
A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su
labwork [276]

Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

    The angle of incidence is  \theta_1 =  10^o

    The refractive index of water is  n_1 = 1.3

  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

4 0
3 years ago
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
Sergio039 [100]
<span>We can use an equation to find the gravitational force exerted on the HST. F = GMm / r^2 G is the gravitational constant M is the mass of the Earth m is the mass of the HST r is the distance to the center of the Earth This force F provides the centripetal force for the HST to move in a circle. The equation we use for circular motion is: F = mv^2 / r m is the mass of the HST v is the tangential speed r is the distance to the center of the Earth Now we can equate these two equations to find v. mv^2 / r = GMm / r^2 v^2 = GM / r v = sqrt{GM / r } v = sqrt{(6.67 x 10^{-11})(5.97 x 10^{24}) / 6,949,000 m} v = 7570 m/s which is equal to 7.570 km/s HST's tangential speed is 7570 m/s or 7.570 km/s</span>
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4 years ago
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