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Anna71 [15]
3 years ago
8

A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s afte

r this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant

Physics
2 answers:
lana66690 [7]3 years ago
6 0

Answer: The ball's acceleration is 2.35 m/s2

Explanation: Please see the attachment below

Luden [163]3 years ago
6 0

Answer:

The acceleration is  a= 2.4 \ m/s^2

Explanation:

From the question we are told that

   The distance covered is  d =  87 \ m

    The time taken is  t =  8.6 \ s

    Time taken reach the bottom is  t_b  = 1 \ s

 According to the equation of motion

           S =  ut  + \frac{1}{2} at^2

since the ball started at rest u =  0 m/s  

     substituting values

    87  =  0 + \frac{1}{2} * a  * (8.6)^2

=>     a =  \frac{2 *  87}{8.6^2}

=>      a= 2.4 \ m/s^2

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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150
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<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

\rho \mathrm{Vg}=2 \mathrm{mg} \\

V=\frac{2 m}{\rho} \\

V=\frac{2 \times 150}{1000} \\

\mathrm{~V}=0.3 \mathrm{~m}^3

Area of pool $=\frac{\pi}{4} d^2$

&=\frac{\pi}{4} \times 6^2 \\

&=28.27 \mathrm{~m}^2

Height of the water rise

h &=\frac{V}{A} \\

h &=\frac{0.3}{28.27} \\

& \mathrm{~h}=0.0106 \mathrm{~m}

  • Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

#SPJ4

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Its B.hope this helps

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