A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
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Consider the diagram below.
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
a)
i) 120 s
ii) 1.57 m/s
b)
i) Refer to the attached diagram
ii) Up
c)
d) Greater than
Explanation:
The problem does not provide full details: consult the attachments for the complete text.
a)
The revolution period of the book equals the total duration needed for the book to make one full revolution.
By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.
We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.
This results in the revolution period being
T = 210 - 90 = 120 s
ii)
The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.
Mathematically:
where
R represents the wheel radius
T = 120 s indicates the period
Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as
d = +30 - (-30) = 60 m
This means the radius calculates to
R = d/2 = 30 m
So, the final speed is
b)
i) Please consult the attached free-body diagram for the book when at its lowest point.
Two forces act on the book at the lowest position:
- The weight of the book, represented as
where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.
- The normal force the bench exerts on the book is represented by N. This force acts upward.
ii)
While at its lowest position, the book maintains a horizontal motion at constant speed.
Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.
According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:
where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.
c)
As discussed in part b), there are two forces influencing the book at the lowest point:
- The weight, , directed downward
- The normal force from the bench, N, directed upward
Given that the book is in uniform circular motion, the net force must match the centripetal force , leading us to the equation:
where
represents the speed of the book
R stands for the radius of the circular path.
We derive an expression for the normal force:
d)
As per the discussions in parts c) and d):
- The normal force acting on the book at its lowest point becomes
- The weight (gravitational force) of the book is
Upon comparing these two equations, we conclude:
Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.
