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sineoko [7]
3 years ago
15

A frog falls from its rainforest tree. If we ignore wind resistance, (a) how much time does it take the frog to fall a distance

of 12.0 m? (b) how fast is the frog falling at this point?
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

Explanation:

a) Using the equation of motion

S = ut + 1/2gt²

S is the distance of fall

g is the acceleration due to gravity

t is the time taken

Given S = 12.0m, g = 9.81m/s^2, un= 0m/s

12 = 0+1/2(9.81)t²

12 = 4.905t²²²

t² = 12/4.905

t² = 2.446

t = √2.446

t = 1.56secs

b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt

v = 0+9.81(1.56)

v = 15.34m/s

Hence the frog is falling at the rate of 15.34m/s

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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
3 years ago
Using the formula 1/2 (m x v2), what is the kinetic energy of a 4 kg rock falling through the air at 5 m/s
aivan3 [116]

Answer:

KE = 50J

Explanation:

KE = \frac{1}{2}mv^{2} \\\\ KE = \frac{1}{2}(4)(5)^{2} \\\\ KE = 2(25) \\\\ KE = 50J

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