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-BARSIC- [3]
3 years ago
13

A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?

Physics
1 answer:
creativ13 [48]3 years ago
8 0

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

s=ut+\dfrac{1}{2}at^2

Herre, u = 0

s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m

So, she will cover a distance of 800 m.

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An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
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a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

7 0
2 years ago
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