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Nataly [62]
3 years ago
8

A particle moves along the x axis. It is intially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.32

0 m/s^2. Suppose it moves as a particle under constant acceleration for 4.50m/s^2. Find its position.
Physics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s  (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

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A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The co
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To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.

Let's start by defining acceleration based on speed and time, that is

a = \frac{v}{t}

On the other hand according to Newton's second law we have to

F=ma

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Replacing the value of acceleration in this equation we have

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Substituting with our values we have

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Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.
natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

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Therefore

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Where total momentum

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In conclusion, centripetal force

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F= 618.9 N

Read more about mass

brainly.com/question/15959704

CQ

Flag

a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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Units of force are the newton pound.
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