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4 years ago

What is the shape of pure carbon fullerenes?

1 answer:

6 0

A fullerene is an allotrope of carbon in the form of a hollow sphere, ellipsoid, tube, and many other shapes. Spherical fullerenes, also referred to as Buckminsterfullerenes or buckyballs, resemble the balls used in association football. Cylindrical fullerenes are also called carbon nanotubes (buckytubes).

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If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, what

DENIUS [597]

Its total mechanical energy is <em>2,000 J</em>.

We don't have enough information to say anything about its heat energy, its chemical energy, or the energy due to any electrical charge it may be carrying or any magnetic field it may have.

6 0

3 years ago

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R. From what mi

Oliga [24]

Answer:

$H = \frac{5}{2}R$

Explanation:

As we know that if the block will complete the circular motion of the path then the speed at the bottom most part of the path must be equal to

$v = \sqrt{5Rg}$

now we know that

velocity at the bottom of the path is due to conversion of potential energy to kinetic energy

so we can say it is given as

$U = KE$

$mgH = \frac{1}{2}mv^2$

now we have

$mgH = \frac{1}{2}m(5Rg)$

$H = \frac{5}{2}R$

3 0

3 years ago

1. As described by the first law of thermodynamics, the change in internal energy of a system is equal to _____.

slava [35]

The first one is D
the second one is C

3 0

3 years ago

Read 2 more answers

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

Describe the circumstances for which the output work would equal the input work in a machine

tiny-mole [99]

That's only possible if there is absolutely zero friction in the machine.

In the real world, we can come close, with lubricants (oil, grease) and
streamlining, but we can never build anything with totally zero loss of
energy inside.

3 0

3 years ago