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3 years ago

7

Infer why gases such as the oxygen used at hospitals are compressed. why must compressed gases be shielded from high temperature

s? what must happen to compressed oxygen before it can be inhaled?

1 answer:

Ipatiy [6.2K]3 years ago

3 0

Compressed gases are highly explosive when exposed to high temperatures, because they expand rapidly.

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Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00

lions [1.4K]

Answer:

b. $\Delta KE = 390 eV$

Explanation:

As we know that the electric field due to infinite line charge is given as

$E =\frac{\lambda}{2\pi \epsilon_0 r}$

here we can find potential difference between two points using the relation

$\Delta V = \int E.dr$

now we have

$\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr$

now we have

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})$

now plug in all values in it

$\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})$

$\Delta V = 216ln6 = 387 V$

now we know by energy conservation

$\Delta KE = q\Delta V$

$\Delta KE = (e)(387V) = 387 eV$

3 0

3 years ago

What did physicist James Joule's famous paddle wheel experiment demonstrate?

PilotLPTM [1.2K]

<u>Answer </u>

A. that the initial gravitational potential energy of the masses transformed into kinetic energy of the paddles and then to thermal energy in the water

<u>Explanation</u>

James Joule allowed some water to fall from a height of 1 foot. the water would turn a paddle wheel at the bottom causing a temperature of water to raise.

The height form which the water fell, mass and the temperature of water was measured and used to calculate mechanical equivalent of heat.

From the choices given the best answer is A. that the initial gravitational potential energy of the masses transformed into kinetic energy of the paddles and then to thermal energy in the water.

7 0

3 years ago

Read 2 more answers

The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u

elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}$

For Part (b)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}$

For Part (c)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}$

8 0

3 years ago

Calculate the frequency of yellow light with a wavelength of 580x10 to the -9

aniked [119]

$5.2x10¹⁴ hz$ is your answer.

3 0

3 years ago

The error in the measurement of the radius of a sphere is 2%. What will be the error in the calculation of its volume?

BlackZzzverrR [31]

To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

$\frac{dr}{r} = \pm 0.02$

The objective is to find the percentage error in the volume.

The volume can be defined as

$V = \frac{4}{3} \pi r^3$

Differentiate with respect the radius we have,

$\frac{dV}{dr} = 4\pi r^2$

$dV = 4\pi r^2 \times dr$

$dV = 4\pi r^2 (\pm 0.02r)$

$dV = \pm 4\times 0.02 \times \pi r^3$

The percentage change in the volume is as follows

$\% change = \frac{dV}{V} \times 100$

$\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100$

$\% change = \pm 6\%$

Therefore the percentage change in volume is $\pm 6\%$

3 0

3 years ago