Answer:
Explanation:
a )
Each blade resembles a rod with its axis positioned near one end.
The moment of inertia for one blade is:
= 1/3 x m l²
where m stands for the mass of the blade
l represents the length of each blade.
Total moment of inertia for 3 blades is:
= 3 x x m l²
ml²
2 )
Details provided include:
m = 5500 kg
l = 45 m
Substituting these values produces:
moment of inertia of one blade:
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia for 3 blades:
= 3 x 37.125 x 10⁵ kg.m²
= 111.375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I denotes the moment of inertia of the turbine
ω symbolizes angular velocity
ω = 2π f
f indicates the rotational frequency of the blades
d )
We have I = 111.375 x 10⁵ kg.m² (Calculated)
f = 11 rpm (revolutions per minute)
= 11 / 60 revolutions per second
ω = 2π f
= 2π x 11 / 60 rad / s
Calculating angular momentum yields
= I x ω
111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹.
Answer:
b. 9.5°C
Explanation:
= Ice mass = 50 g
= Initial temperature of water and aluminum = 30°C
= Latent heat of fusion =
= Water mass = 200 g
= Water's specific heat = 4186 J/kg⋅°C
= Aluminum mass = 80 g
= Aluminum's specific heat = 900 J/kg⋅°C
The equation governing heat exchange in the system is presented by
The final equilibrium temperature calculates to 9.50022°C
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.
Answer:
Part A: 7.75 m/s
Part B: 2330.8 kN
Part C: 24.03 kN
Part D: 4.8 kN
Part E:
Part F: Option D
Bending one's legs lengthens the duration of force application from the ground, resulting in a reduction of the applied force.
Explanation:
Part A
Using the fundamental kinematic equations
where v represents the velocity just before ground impact, g denotes gravitational acceleration, u signifies initial velocity, and h is the fall height.
With the initial velocity at zero, thus:
Plugging in 10 m/s² for g and 3 m for h gives:
Part B
The force exercised by the leg can be expressed as
F = PA where P is pressure, F indicates force, and A denotes the cross-sectional area of the bone.
With a substitution of 2.3 cm or 0.023m for d and for P, we derive the force as:
Part C
The fundamental kinematic equations from part (a) can also be rearranged to show:
and solving for a yields
where a is the acceleration and
signifies the change in length.
Using the previously derived value from part a, 7.75 m/s for v, and 0.01 m for gives us:
The force felt by the man is given by:
Part D
A similar approach with the fundamental kinematic equations shows:
and solving for a indicates:
where a is the acceleration and
denotes the change in height.
The force experienced can thus be defined as .
For substitution, we use m = 80 Kg, and 0.5m for \triangle h along with other values calculated in part c.
Part E
Part F
Bending one's legs extends the period over which the force acts, thus lessening the overall force exerted by the ground.