Please help me i have this due tommorrow!!!

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago

A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran

dedylja [7]

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= $I^{2}R t$

let R * t = k ( ∴ R and t both are constant)

so E= k $I^{2}$

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k $I^{2}$

Cancel same terms on both sides.

I₂^2= 4* $I^{2}$

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

3 0

3 years ago

Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1

anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

ΔL/Lo = αΔT

solve for ΔL

ΔL = αLoΔT

Substitution

ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

ΔL = 0054 m = 5.4 cm

7 0

3 years ago

It says find the slope for each line I'm stuck on number one can you help me

Allushta [10]

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

(-2, -1)(3, 1)

Therefore,

$\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}$

7 0

1 year ago

8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

3 years ago