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:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
=
Therefore the gravitational force is halved. That is it will always decrease
Answer:
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The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>