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Papessa [141]
2 years ago
11

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 9.8 s later its speed was 20 m/s in the oppo

site direction. What was the average acceleration of the particle during this 9.8 s interval
Physics
1 answer:
dem82 [27]2 years ago
6 0

The average acceleration of the particle in the time interval of 9.8 s is 10.204 m/s² opposite to the direction of motion.

Given:

Speed of particle, v₁ = 80 m/s (in positive x-direction)

Speed of particle, v₂ = -20 m/s (in opposite direction)

Time interval, Δt = 9.8 s

Calculation:

We know that, the average acceleration is given as:

a_avg = (v₂ - v₁)/ Δt     - ( 1 )

Applying values in above equation we get:

a_avg = (v₂ - v₁)/ Δt

           = (-20 m/s - 80 m/s) / (9.8 s)

           = -10.204 m/s²

Therefore the average acceleration of the particle in the time interval Δt is 10.204 m/s² opposite to the direction of motion.

Learn more about average acceleration here:

<u>brainly.com/question/27184422</u>

#SPJ4

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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
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Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

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x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

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\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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<h3>Further explanation </h3>

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Keywords: inertia, Newton's First Law

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