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irina1246 [14]
3 years ago
7

A ball rolls from x = - 5m to x = 0m in 1 second . What was its average velocity? (Units = m/s) Don't forget: velocities and dis

placements to the right are +, to the left are -
Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

+ 5 m/s

Explanation:

change in displacement = ΔX=final position - initial position

                                            ΔX = 0-(-5) =0+5 =+ 5 m

average velocity =  ΔX/t

                             = +5/1

                             = + 5 m/s

positive sign shows that ball rolls towards right

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vladimir1956 [14]

Answer:

3600N

Explanation:

F=m\frac{v^2}{r}

4 0
2 years ago
1. George is traveling to Boston from Springfield. Springfield is 144
denpristay [2]

Answer:

48 kilometers per hour

Explanation:

144/3 = 48 km/h

3 0
2 years ago
I have no clue what to do please help.
Yakvenalex [24]

Answer:

Step one : read the directions

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Explanation:

3 0
2 years ago
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
a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being
marin [14]

From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

  35 ft + 27 ft + 27 ft + 18 ft = 107 ft

3 0
3 years ago
Read 2 more answers
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