¡Hellow!
For this problem, first, lets convert the seconds in hours:
5,4x10³ 5400
h = sec / 3600
h = 5400 s / 3600
h = 1,5
Let's recabe information:
d (Distance) = 386 km
t (Time) = 1,5 h
v (Velocity) = ?
For calculate velocity, let's applicate formula:
Reeplace according we information:
386 km = v * 1,5 h
v = 386 km / 1,5 h
v = 257,33 km/h
The velocity of the train is of <u>257,33 kilometers for hour.</u>
<u></u>
Extra:
For convert km/h to m/s, we divide the velocity of km/h for 3,6:
m/s = km/h / 3,6
Let's reeplace:
m/s = 257,33 km/h / 3,6
m/s = 71,48
¿Good Luck?
Congress has the ability to override a veto by a two-thirds majority vote.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
learn more about oscillators brainly.com/question/15169199
#SPJ1
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470