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Romashka [77]
3 years ago
14

A hedgehog runs 500 meters in 10 minutes, then 300 meters in 8 minutes and finally 100 meters in 6 minutes. What is the hedgehog

's average speed in meters per minutes?
Physics
1 answer:
Marina86 [1]3 years ago
5 0

200 meters per minutes

500 meters -300 meters= 200 meters

300 meters -100 meters =200 meters

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A _____ is a quantity that has magnitude and direction
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vector is the answer of this blank

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Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
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Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

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              y = v₀ t + ½ g t²

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Acceleration changes and objects...
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Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses
Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity  (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt  

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:  

h: hight in meters (m)    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁  :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g      Equation (1)

vf₁²= v₀₁²-2*g*h₁   : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g)   Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂²  v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂²   Equation (3)

Equation that relates h₁ to h₂

h₂=  h₁ + 56.3  ,  h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3  Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0
3 years ago
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