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Sergio039 [100]
3 years ago
15

Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

ose meg wants to apply the scientific method to discover reasons for the hot pavement. What is the next step she should take?
Physics
1 answer:
mart [117]3 years ago
7 0
1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.
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A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be
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Answer:

h = 10.4 m

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v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t  :

Equations of the uniformly accelerated rectilinear motion of upward   (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m  

y: vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j  ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the  initial  vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 =  (v₀y)²

v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }

v₀y = 14.3 m/s

Calculation of the maximum height  the ball rise (h)

In the maximum height vfy=0

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(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

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vfy = v₀y -gt

0 = v₀y -gt

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t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

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We replace data in the equation (1)

x =vx*t    vx= 7.7 m/s , t =2.92 s  (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

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\alpha = tan^{-1} (\frac{-14.3 }{7.7 })

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

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