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Anna007 [38]
3 years ago
5

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a

tmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Physics
1 answer:
s2008m [1.1K]3 years ago
6 0

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

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Help with this question plz and show how you got it :) tysm!
spayn [35]
The hardest part of the job is to find the right formula to use, and write it down. You've already done that ! The rest is just turning the crank until an answer falls out.

You wrote. E = m g h.
Beautiful.
Now divide each side by (g h), and you'll have the formula for mass:

m = E / (g h).

You know all the numbers on the right side. Just pluggum in, do the arithmetic, and you'll have the mass.
5 0
3 years ago
Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work
LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

\frac{\Delta x}{\Delta t}

Analysing the units, meters (displacement) and seconds (time) are basic units:

\frac{m}{s}

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

F = ma

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

a=\frac{\Delta v}{\Delta t}

Analysing the units:

\frac{\frac{m}{s}}{s}=\frac{m}{s^2}

Therefore, the unit of force is:

kg\frac{m}{s^2}

Pressure:

Pressure is given by the equation:

P=\frac{F}{S} where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

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Dividing the aforementioned unit of force by the unit of area:

\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}, the unit of pressure

Work:

Work is given by the equation:

W=\vec{F}\cdot \vec{x}, (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}, the unit of work.

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Answer:a, d, e

Explanation:

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I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
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