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Snowcat [4.5K]
1 year ago
5

Maria throws two stones from the top edge of a building with a speed of 20 m/s. She throws one straight down and the other strai

ght up. The first one hits the street in a time t1. How much later is it before the second stone hits
Physics
1 answer:
Georgia [21]1 year ago
7 0

Maria throws two stones from the top edge of a building at a speed of 20 m/s, The time is mathematically given as

t=4.5 sec.

<h3>How much later is it before the second stone hits?</h3>

Generally, the equation for speed  is mathematically given as

V = u+at

Therefore

0 = 22-9.8xt.

t = 22 /9.8 sec

t = 2.24489.

The same time will be taken from max^m height to point O

The extra time is taken

t=4.49 sec.

t=4.5 sec.

Read more about Time

brainly.com/question/4931057

#SPJ1

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) The centre of a mass, m, is at a distance, r, from the centre of a larger mass M. Assuming there are no other masses present,
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You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 m in front of you. your reaction
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The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a

x₂ = 0.5*a*t² = 0.5*v°²/a

The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7

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