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OleMash [197]
3 years ago
8

Please help asap How can carbon monoxide be properly disposed? if you are right you will get brainliest

Physics
1 answer:
Elis [28]3 years ago
6 0
The best way to treat CO poisoning is to breathe in pure oxygen. This treatment increases oxygen levels in the blood and helps to remove CO from the blood. Your doctor will place an oxygen mask over your nose and mouth and ask you to inhale.

Is that what you mean? Treat the poisoning? It’s a gas, so you can dispose of the gas. Unless you mean dispose of the detector?
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What statement BEST describes a solid?
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4 years ago
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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
Nesterboy [21]

Answer:

Compression distance: d \approx 0.102\,m

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

K + W_{l} = U (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2} (2)

Where:

m - Mass of the block, in kilograms.

v - Final velocity of the block, in meters per second.

\mu - KInetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s - Width of the rough patch, in meters.

k - Spring constant, in newtons per meter.

d - Compression distance, in meters.

If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2}

m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

d \approx 0.102\,m

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It must be either speeding up, or slowing down, or turning. There are no other possibilities.

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