Energy is observed in two basic forms: potential and kinetic. Which of the following correctly matches these forms with a source
1 answer:
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Answer:
La magnitud del momento total es de 21.2 kg m/s y su dirección es de 39.5° respecto al eje x.
Explanation:
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El momento total se calcula como la suma de los momentos de las piezas.
El momento de cada pieza se calcula de la siguiente manera:
p = m · v
Donde:
p = momento.
m = masa.
v = velocidad.
El momento es un vector. La pieza de 200 g se mueve a lo largo del eje x, por lo que su momento será:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
La pieza de 300 g se mueve a lo largo del eje y. Su vector momento será:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
El momento total es la suma de cada momento:
Momento total = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Momento total = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Momento total = (16.4 kg m/s, 13.5 kg m/s)
La magnitud del momento total se calcula de la siguiente manera:
La dirección del vector de momento se calcula utilizando trigonometría:
cos θ = px/p
Donde px es el componente horizontal del momento total y p es la magnitud del momento total.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° si no redondeamos la magnitud del momento total)
<pFinalmente, la magnitud del momento total es 21.2 kg m/s y su dirección es 39.5° respecto al eje x.
Answer:
176.38 rpm
Explanation:
The proportion of mass for arms and legs is 13%.
For legs and trunk, it's 80% of the total mass.
Additionally, the head accounts for 7% of the total mass.
Overall, the mass of the skater is 74.0 kg.
Each arm length is 70 cm, or 0.7 m.
The skater's height is 1.8 m, and the trunk diameter measures 35 cm, or 0.35 m.
Starting angular momentum is 68 rpm.
We make the following assumptions:
- The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
- Friction between the skater and the ice is considered negligible.
To analyze her body, we divide it into two parts, treating the arms as spinning rods.
1. Each arm (rod) has a moment of inertia of .
The arms comprise 13% of 74 kg, calculated as 0.13 x 74 = 9.62 kg.
Each arm is then evaluated as 9.62/2 = 4.81 kg.
Let L represent the length of each arm.
Thus,
I = x 4.81 x
= 0.79 kg-m for each arm.
2. The body, treated as a cylinder, has a moment of inertia of .
For the body, the radius r is half of the trunk diameter: r = 0.35/2 = 0.175 m.
The mass of the trunk amounts to (80% + 7%) of 74 kg, which calculates to 0.87 x 74 = 64.38 kg.
Therefore, I = x 64.38 x
= 0.99 kg-m.
Two cases are considered:
case 1: Body spinning with arms extended.
Total moment of inertia equals the combined moments of inertia of both arms and the trunk.
I = (0.79 x 2) + 0.99 = 2.57 kg-m.
The angular momentum is given by Iω.
Here, ω = angular speed = 68.0 rpm = x 68 = 7.12 rad/s.
The angular momentum then becomes 2.57 x 7.12 = 18.29 kg-rad/m-s.
case 2: Arms drawn in alongside the trunk.
The moment of inertia is attributed solely to the trunk. This is 0.91 kg-m.
The angular momentum equals Iω.
= 0.99 x ω = 0.91ω.
By the principle of conservation of angular momentum, the two angular momentum quantities are equal. Therefore,
18.29 = 0.99ω.
Solving gives ω = 18.29/0.99 = 18.47 rad/s.
This leads to 18.47 ÷ = 176.38 rpm.
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.
The height from which the drops fall is 3.57m
Assuming the drops fall at a rate of 1 drop every t seconds. The initial drop takes 5t seconds to hit the ground. The second drop reaches the bottom of the 1.00 m window in 4t seconds, while the third drop reaches the top of the window in 3t seconds.
Calculate the distances covered by the second and third drops s₂ and s₃, which begin from rest at the building's roof.
The height of the window s is described by,
The first drop has reached the base after taking 5t seconds.
The roof height h represents the distance covered by the first drop and is expressed as,
the height of the roof remains at 3.57 m
