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Vera_Pavlovna [14]
4 years ago
11

Which environmental changes occur faster?

Physics
2 answers:
irga5000 [103]4 years ago
6 0
The answer is B. Oil Spill
enot [183]4 years ago
4 0

He/She is right!

Yes. I just did the test the answer is

B). Oil Spill

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State two condition necessary for a solid to float in a liquid
hoa [83]

Answer:

it's density must be less than water

law of floatation

wt of the immerged body = wt of the water displaced by it

4 0
3 years ago
If the mass of a material is 41 grams and the volume of the material is 5 cm^3, what would the density of the material be? answe
Mila [183]
Density can be calculated using the following rule:
Density = mass/volume

We are given that:
mass = 41 grams
volume = 5 cm^3

Substitute with the givens in the above equation to get the density as follows:
Density = 41/5 = 8.2 g/cm^3

Based on the above calculations:
answer : 8.2
units : g/cm^3
6 0
3 years ago
Read 2 more answers
Melissa uses bags of mulch that way 40 pounds each. She needs to carry 10 bags to her truck. She usually carries one bag at a ti
arsen [322]
No. She would be doing the same amount of work that way. Work is defined to be equal to the force multiplied by the distance. Carrying two bags at a time would cause her to exert twice the effort, so the total amount of work done in the end would still be the same.
4 0
3 years ago
Identify and explain one way in which we try to deal with infiltration? What are the benefits and cost of that particular soluti
tigry1 [53]

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i vigilate

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5 0
3 years ago
Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
koban [17]

Answer:

A

   N  = 1340.86 \ slits  / cm

B

    \theta  = 15.7^o

Explanation:

From the question we are told that  

      The wavelength is  \lambda  =  650 \  nm  =  650  *10^{-9} \  m  

        The angle of  first bright fringe is  \theta  =  5^o  

        The order of the fringe considered is  n  =1

Generally the condition for constructive interference is  

       dsin (\theta ) = n * \lambda

=>    d =  \frac{1 *  650 *10^{-9 }}{ sin(5)}

=>    d = 7.458 *10^{-6} \  m

Converting to cm

           d = 7.458 *10^{-6} \  m = 7.458 *10^{-6}  * 100 =  0.0007458 \  cm

Generally the number of grating pre centimeter is  mathematically represented as

           N  =  \frac{1}{d}

=>         N  =  \frac{1}{0.0007458}

=>         N  = 1340.86 \ slits  / cm

Considering question B  

   From the question we are told that

     The first wavelength is  \lambda_1 =  650 \ nm  =  650 *10^{-9} \  m

     The second wavelength is  \lambda_2 = 429 \  m   =   420 *10^{-9 } \  m

      The order of the fringe is  n  =  2

       The grating is  N =  5000 \  slits / cm

Generally the slit width is mathematically represented as

              d =  \frac{1}{N  }

=>          d =  \frac{1}{ 5000  }

=>          d =   0.0002 \  c m  =  2.0 *10^{-6} \ m

Generally the condition for constructive interference for the first ray is mathematically represented as

         d sin(\theta_1) =  n *  \lambda_1

=>      \theta_1 = sin^{-1} [\frac{ 2 *  \lambda }{d}]

=>       \theta_1 = sin^{-1} [\frac{ 2 *   650 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_1 = 40.5 ^o

Generally the condition for constructive interference for the second ray is mathematically represented as

         d sin(\theta_2) =  n *  \lambda_2

=>      \theta_2 = sin^{-1} [\frac{ 2 *  \lambda_1 }{d}]

=>       \theta_2 = sin^{-1} [\frac{ 2 *   420 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_2 = 24.8  ^o

Generally the angular separation is mathematically represented as

            \theta  =  \theta_1 - \theta_1

=>          \theta  = 42.5^o -  24.8^o

=>          \theta  = 15.7^o

4 0
3 years ago
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