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dexar [7]
3 years ago
10

a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

e origin with a velocity of -12 m/s, what js its position at t=4.0s?
Physics
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

 Position at t= 4 seconds is 144 m

Explanation:

 It is given that acceleration, a = 18 t, where t is the time.

 We know that Velocity, v = \int { a} \, dt

  Substituting value of a,

           Velocity, v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c

 We know that at t = 0, v = -12 m/s

         So, 9*0^2+c=-12\\ \\ c=-12m/s

So velocity, v = (9t^2-12)m/s

  We also know that displacement, x = \int { v} \, dt

     Substituting value of v,  

        Displacement, x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c

  We know that at t = 0, particle is at origin, x =0.

               So,  0=3*0^3-12*0+c\\ \\ c=0

   Displacement, x = 3t^3-12t

At t = 4 seconds

   x = 3*4^3-12*4=192--48=144m

Position at t= 4 seconds is 144 m  

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The lightbulb that carries more current will be the 25W bulb.

<h3>How to explain the information?</h3>

It should be noted that an electric current simply means the stream of charged particles that move through an electrical conductor or space.

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You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

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

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
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Answer:

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Explanation:

earth's plates move only a few centimeters per year.

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