Wellll . . . it's doubtful that you'd hear the sound of a train
from 100 km away (about 62 miles), but this is a fun problem
so let's go through it just for the math.
Online, I looked up the speed of sound in various materials.
Here's what I found:
-- Speed of sound in normal air . . . . . 343 m/s
-- Speed of sound in iron . . . . . 5,130 m/s
But let's go a little farther !
Rails used to be made of cast iron or wrought iron.
But now they're made of hot rolled steel.
-- Speed of sound in steel . . . . . 6,100 m/s
Time to cover the distance = (distance) / (speed)
Time through air = (100,000 m) / (343 m/s)
= 291.5 sec = 4 minutes 51.5 seconds
Time through old iron rails = (100,000 m) / (5,130 m/s)
= 19.5 seconds (272 sec sooner)
Time through new steel rails = (100,000 m) / (6,100 m/s)
= 16.4 seconds (275 sec sooner)
Answer:
The magnitude of the electric field between the plates is half its initial value.
Explanation:
We know the electric field E = V/d where V = voltage applied and d = separation between plates.
Since V is constant and V = Ed,
So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.
So, E₂ = E₁d₁/d₂
Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁
So, E₂ = E₁d₁/2d₁ = E₁/2
So, E₂ = E₁/2
Thus, the magnitude of the electric field between the plates is half its initial value.
Answer:
current moves from positive terminal of a cell towards the negative terminal of a cell.
Explanation:
this happens when good conductors which have free electrons moving randomly within them.when a cell i s connected across the ends the free electrons moves in a given direction. then the current starts flowing in a opposite direction with electrons which moves negative terminal to positive terminal.
Answer:
–2.25 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 15 m/s
Distance travelled (s) = 50 m
Final velocity (v) = 0 m/s
Deceleration (a) =?
v² = u² + 2as
0² = 15² + (2 × a × 50)
0 = 225 + 100a
Collect like terms
0 – 225 = 100a
– 225 = 100a
Divide both side by 100
a = –225/100
a = –2.25 m/s²
Thus, the deceleration of the vehicle is –2.25 m/s²
