All of these are types of electromagnetic waves except for:

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

3 years ago

Question 15 of 15

castortr0y [4]

Answer:

a and b

Explanation:

are the correct answers

7 0

3 years ago

What is the equation for elastic potential energy?

drek231 [11]

Answer: HOPE THIS HELPED YOU! :D ;P

Elastic potential energy = force x distance of displacement.

Explanation:

6 0

3 years ago

Read 2 more answers

The platofm on the school stage is 8 3/4 feet wide. each chair is 1 5/12 feet wide. how many chairs will fit acroos the platform

wariber [46]

6 chairs will fit across the platform.

<h3>Calculation</h3>

We have a school stage of 8 3/4 feet wide.

It's also given that each chair is 1 5/12 feet wide.

To find out the number of chairs possible we can simply divide the total length available by width of the chair.

8 3/4 = 35/4

1 5/12 =17/12

So, 35/4 / 17/12

So, the final answer comes up to 6.

so, the total number of chairs that would fit in are 6.

To know more about random number intervals, visit:

brainly.com/question/13657605

#SPJ4

3 0

1 year ago

A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces

schepotkina [342]

Explanation:

(a) Formula to calculate volume of the submerged wooden block is as follows.

$V_{sub} = l \times w \times d$

It is given data of the wooden block is as follows.

depth = 7.96 cm, length (l) = 6 cm

width (w) = 4 cm,

So, we will calculate the volume of the submerged wooden block as follows.

$V_{sub} = l \times w \times d$

= $6 \times 6 \times 7.96$

= 286.56 $cm^{3}$

Hence, the submerged volume of the block is 286.56 $cm^{3}$ .

(b) Expression for the buoyant force acting on the wooden block is as follows.

$F_{B} = \rho_{w} g V_{sub}$

And, expression for the force of gravity of the wooden block is as follows.

$F_{g} = m_{b}g$

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

$F_{g} = F_{B}$

Hence, mass of the wooden block will be calculated as follows.

$F_{g} = F_{B}$

$m_{b}g = \rho_{w}gV_{sub}$

$m_{b} = \rho_{w}V_{sub}$

= $997 kg/m^{3} \times 286.56 cm^{3}$

= $997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}$

= 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c) Expression for the density of the block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

Now, expression for the total volume of the wooden block is as follows.

$V_{b} = l \times w \times h$

Hence, density of the given block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

= $\frac{m_{b}}{lwh}$

= $\frac{0.02857 kg}{4 \times 4 \times 15}$

= $1.19 \times 10^{-4} kg/cm^{3}$

Therefore, density of the given block is $1.19 \times 10^{-4} kg/cm^{3}$ .

3 0

3 years ago