Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6
- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6 *(15s)²
Solving:
x=½*6 *(15s)²
x=½*6 *225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>
Answer:
depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
Answer:
Hey
Yes, this is true.
As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.
States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.
