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
<span>C) mountains; two continental plates meet at a convergent boundary</span>
Answer:
a. Your mass with the basketball is 55.5 kg
b. Your new velocity is 0.045 m/s
Explanation:
We first add your weight and the basketballs weight to get 55.5 kg.
Then to find b. we use the equation: v final = (m1 * v1) / (m1 +m2)
So m1 is the basketball which is 0.5 kg and v1 is 5 m/s. So the top half is (0.5 * 5)
The bottom half is just our weights added together.
Answer:
d₁ = 0.29 in
d₂ = 0.505 in
Explanation:
Given:
T = 1500 lbf in
L = 10 in
x = 0.5 L = 5 in
First case: T = T₁ + T₂
T₂ = T - T₁ = 1500 - 750 = 750 lbf in
If the shafts are in series:
θ = θ₁ + θ₂
θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)
Second case: If d₁ ≠ d₂
θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)
t₁ = t₂
(eq. 2)
T₁ + T₂ = 1500 (eq. 3)
θ₁ first case = θ₁ second case
Replacing:
The same way to θ₂:
From equation 2, we have:
d₁ = 0.587 * d₂
From equation 3, we have:
d₂ = 0.505 in
d₁ = 0.29 in
Answer:
Phosphorus
Explanation:
Phosphorus needs 3 electrons to complete its outter most shell of electrons
