¡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>
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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?
In this question, you're determining the time (t) taken for an object to fall from a distance (d).
The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)
d = 8,848m and g = 9.8m/s
Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)
The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)
Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)
Now for the last step, find the square root of the remaining number:
t = 42.5s
So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.
I hope this helps :)
Accelerated motion such as a vehicle (car) or a moving item such as a (football thrown in the air)
Answer:
(a) A. Uniform line of charge and B. Uniformly charged sphere
(b) To three digits of precision:
λ = 1.50 * 10^-10 C/m
p = 2.81 * 10^-4 C/m^3
Explanation:
