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uysha [10]
3 years ago
15

A ledge on a building is 20 m above the ground. A taut rope attached to a 4.0 kg can of paint sitting on the ledge passes up ove

r a pulley and straight down to a 3.0 kg can of nails on the ground. If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?
Physics
1 answer:
Andrews [41]3 years ago
8 0

Answer:

 t = 5.4 s

Explanation:

from the question we are given :

height (s) = 20 m

mass of paint (Mp) = 4 kg

mass of nails (Mn) = 3 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

  • The net force accelerating the can of paint should be equal to the difference in weight of the can of paint and the can of nails.

            weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N

            weight of paint = mass nails x g = 4 x 9.8 = 39.2 N

             net force = 39.2 - 29.4 = 9.8 N

  • net force = total mass x acceleration

             9.8 = (3 +4) x a

              a = 1.4 m/s^{2}

  • from S = Ut + 0.5at^{2}  we can get  the time the carpenter has to catch the nails

          where U is the initial velocity and is 0 since the can was initially at            

            rest

           20 = (0 x t) + (0.5 x 1.4 x t^{2})

            20 = 0.7 x t^{2}

             t^{2} = 28.6

             t = 5.4 s

                   

         

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The  mass of the large truck is determined as 5680 kg.

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The mass of the truck is calculated as follows;

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If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
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0.82 mm

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substituting the parameters in the above equation; we have:

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The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

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\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

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Answer:

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