2 j x b
H c e c hope this helppppsssssssss
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
A material that can easily flow is called a...
A. Fluid.
Answer:
94
Explanation:
f = 2.57 x 10^13 Hz
E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J
Energy of each photon = h f
Where, h is Plank's constant
Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J
Number of photons = Total energy / energy of one photon
N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94