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Misha Larkins [42]
3 years ago
12

At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the oppo

site direction. What is the average acceleration of the particle during this 2.4 s interval?
Physics
1 answer:
hoa [83]3 years ago
3 0
<h2>Average acceleration of the particle during the 2.4 s interval is -20 m/s²</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u =  18 m/s

     Final velocity, v =  -30 m/s    

     Time, t = 2.4 s

     Substituting

                      v = u + at  

                      -30 = 18 + a x 2.4

                      a = -20 m/s²

     Acceleration is -20 m/s²

Average acceleration of the particle during the 2.4 s interval is -20 m/s²

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A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line
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Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

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A 3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-
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3 years ago
1) Write Newton's three laws of motion (20 points). 2) Define acceleration (20 points). 3) Write the Month and Day of the Assign
vaieri [72.5K]

Answer:

Hey

I have no idea when YOUR assignment is due.

Newtons 1rst law:

An object that has constant motion will remain at that speed unless acted on by an external force.

Newtons 2nd law:

F=ma (force=mass*acceleration)

Newtons 3rd law:

when a force is applied to an object, there will be an opposite but equal reaction.

Acceleration:

How much your speed increases/decreases per unit of time.

I wrote all that^

4 0
2 years ago
A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire
EleoNora [17]

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec

Angular frequency is equal to \omega =\sqrt{\frac{k}{m}}

10.362 =\sqrt{\frac{k}{0.041}}

Squaring both side

107.371 ={\frac{k}{0.041}}

k = 4.40 N/m

For vertical osculation

mg=kA

0.041\times 9.8=4.40\times A

A = 0.091 m

So amplitude will be equal to 0.0391 m

8 0
3 years ago
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