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melisa1 [442]
2 years ago
10

Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coef

ficient of kinetic friction between the player and the ground is 0.70. His speed at the start of the slide is 8.23 m/s. I a) Calculate his acceleration during the slide. b) How long (in time) does he slide until he stops?​
Physics
1 answer:
Alex777 [14]2 years ago
4 0

A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>
  • Mass (m) = 81.5 Kg
  • Coefficient of friction (μ) = 0.7
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
  • Frictional force (F) =?

F = μN

F = 0.7 × 798.7

F = 559.09 N

<h3>A. How to determine the acceleration</h3>
  • Mass (m) = 81.5 Kg
  • Frictional force (F) = 559.09 N
  • Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>
  • Initial velocity (u) = 8.23 m/s
  • Final velocity (v) = 0 m/s
  • Decceleration (a) = -6.86 m/s²
  • Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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Answer:

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Explanation:

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Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

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y = √0.75

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6 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

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Explanation:

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mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

v   =\sqrt{\frac{7.2}{200*10^{-3}}}

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

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A force of 200N acts between two objects at a certain distance apart .the value of the force when the distance is halved is?
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Answer:

800N is the answer

Explanation:

hope it helps

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3 years ago
Aunt Jane weights 45 Newtons. What is her mass?
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Answer:

10.116 Pounds/45 newtons = 10.1164024 pounds/force

Explanation:

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