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3 years ago

Determine the rotational inertia of the construction about an axis perpendicular to the picture and passing through point B.

Physics

1 answer:

sasho [114]3 years ago

6 0

Answer:

m* 8.0 kg*m²

Explanation:

Rotational inertia of a set of discrete masses (considered point masses), regarding an axis passing through a given point, is just the sum of the products of each mass times the distance to the point of interest.

In the attached picture, neglecting the masses of the rods joining the masses, and taking into account that all masses are equal each other, being equal to m kg, we can write the following equation:

I = m₁*r₁² + m₂*r₂² + m₃*r₃² + m₄*r₄²,

⇒ I = m kg*0 +m kg*2 m² +m kg*2m² + mkg*4m²

⇒ I = m*8 kg*m²

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Sound waves are longitudal waves meaning they go back and forth

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3 years ago

Draw/Label a motor OR give directions in words on how to build a motor.

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3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

4 years ago

Does the air exert a buoyant force on all objects in air or only on objects such as balloons that are very light for their size?

Citrus2011 [14]

Answer:

See explanation

Explanation:

Solution:-

Buoyancy is the force that causes objects to float. It is the force exerted on an object that is partly or wholly immersed in a fluid. Buoyancy is caused by the differences in pressure acting on opposite sides of an object immersed in a static fluid. It is also known as the buoyant force. Buoyancy is the phenomena due to Buoyant Force.

It is as an upward force exerted by a fluid that opposes the weight of an object immersed in a fluid. As we know, the pressure in a fluid column increases with depth. Thus, the pressure at the bottom of an object submerged in the fluid is greater than that at the top. The difference in this pressure results in a net upward force on the object which we define as buoyancy.

- The formula for buoyant force (Fb) is given:

Fb = ρ*g*V

- The force acts on all objects. However, it depends on the fluid density and amount of volume displaced.

- The Buoyant force exerted by air with density = 1.225 kg/m^3 on an object with volume (V) is:

Fb = ρ*g*V = 1.225*9.81*V = 12.02*V

- For the similar object with mass (m), the downward weight would be:

W = m*g

- For the object to float the buoyant force (Fb) must be greater than weight of the object:

Fb > W

12.02*V > m*9.81

V / m > 0.816

- The ratio of V / m must be at-least = 0.816.

- Assuming the object is fully immersed in air, then the volume displaced V = ρ_material*V

ρ_material < 1 / 0.816

ρ_material < 1.225 or ( ρ_air )

- So the for an object to float in air its material density must always be less than that of air. That why in balloons lighter gas is used which have density less than that of air like Helium.

4 0

4 years ago

A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}$

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.

3 0

3 years ago