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marysya [2.9K]
3 years ago
10

Many types of scientific equipment are used to perform different functions in the science lab.Which of the folllowing combinatio

ns of equipment would be needed to bring one liter of water to 85 degree
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinations of equipment would be needed to bring one liter of water to 85°C? a. ... Various pieces of safety equipment are used in the lab to provide protection against injury.

Explanation:

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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

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what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

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