- neon signs
- fireworks
- rockets during the ascent
- meteors
- fireflies
- stars
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.
Answer:
Billions :)
<em>hope this helps! <3</em>
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
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