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igor_vitrenko [27]
3 years ago
10

What happens to the heat energy the particles receive during convection and conduction?

Physics
1 answer:
kherson [118]3 years ago
4 0
The particals speed up
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A motorcycle begins at rest and accelerates uniformly S7.9 we want to find a time to take the motorcycle to reach a speed of 100
Len [333]

The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

v = 100 km/h = 27.8 m/s is the final velocity

a=7.9 m/s^2 is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s

Learn more about accelerated motion:

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6 0
3 years ago
An incandescent light bulb produces light when electrons flow through the
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<span>Electric current passes through a filament of an incandescent bulb, thereby increasing it temperature. When current flows, it contains electrons through the filament to produce light. The answer is c. Typically, incandescent light bulb consists of a glass enclosure that contains tungsten filament. The glass enclosure contains either a vacuum or an inert gas that serves as the filament protection from evaporating. Incandescent light bulbs contain a stem attached at to its base to allow the electrical contacts to run through the envelope without gas or air leaks</span>
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3 years ago
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Which of the following is not an example of work, according to the scientific definition
bogdanovich [222]

Leaning against a brick wall.

All the others use scientific forces of work.

-Steel jelly.

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Convert 3.45inches into km
Katyanochek1 [597]

Answer:

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Explanation:

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2 years ago
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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
2 years ago
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