Answer:
a. 
b. 
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c. 
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
- speed of lagging car,
- distance between the cars,
- deceleration of the leading car after braking,
a.
Time taken by the car to stop:
where:
, final velocity after braking
time taken
b.
using the eq. of motion for the given condition:
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
is the time taken to stop after braking
Answer:
B: Horizontally to the left
Explanation:
Horizontal velocity is always constant throughout the entire trajectory of the rocket and acts in the horizontal direction in which the rocket was launched. This is because gravity only acts in the downwards vertical direction.
So in order words at peak height, horizontal velocity is in the horizontal direction in which the rocket was launched.
So if it was to the left, then direction is left but if right, then direction is right.
Looking at the options, the most appropriate will be:
Horizontally to the left
Answer:
your question in not complete.
you need to the high too.
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω
= (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω
= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω + αt
α= (ω - ω )/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ
= ω t + 1/2αt²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω + αt
t= (ω - ω)/α => (0- 3.333)/-1.25
t= 2.67s
