Solve the questions in the picture

Given a manhole with the outlet invert elevation of 2605. A run of 48 inch pipe connects to the next manhole at an invert elevat

alexdok [17]

Answer:

Distance between manholes = 166.67 (Unit) (Approx)

Explanation:

Given:

Invert elevation = 2605

New Invert elevation = 2610

Grade % = 3% = 0.03

Find:

Distance between manholes

Computation:

Grade % = (Change in elevation)/Distance between objects

0.03 = (2610-2605)/Distance between manholes

0.03 = 5 / Distance between manholes

Distance between manholes = 166.67 (Unit) (Approx)

5 0

3 years ago

Why do we feel chilled in the winter but comfortable in the summer under conditions for which the same room temperature is maint

pantera1 [17]

Answer:

Explanation:

We feel chilled in summer but comfortable in the summer under the same temperature because of our thermal comfort.

Thermal comfort is a function of heat loss from the body and during winter there is excessive heat loss due to heat flux.

As room temperature is fixed so summer and winter comfort cannot be attributed due to heat loss by convection. Due to differences in heat flux radiations in summer and winter there is a chilled feeling associated with winters.

4 0

4 years ago

The pressure of a gas in a rigid container is 125kpa at 300k, what we be the new pressure if the temperature increases to 900k

kipiarov [429]

Answer:

375 KPa

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 125 KPa

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 900 K

Final pressure (P₂) =?

The new (i.e final) pressure of the gas can be obtained as follow:

P₁/T₁ = P₂/T₂

125 / 300 = P₂ / 900

Cross multiply

300 × P₂ = 125 × 900

300 × P₂ = 112500

Divide both side by 300

P₂ = 112500 / 300

P₂ = 375 KPa

Thus, the new pressure of the gas is 375 KPa

7 0

3 years ago

Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg

babunello [35]

Answer:

0.

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

$\sigma_A = -4.10^{-5}C/m^2$

$\sigma_B= -7.10^{-5}C/m^2$

$\sigma_C = -3.1^{-5}C/m^2$

We find the electric field for each one, this formula is given by,

$E= \frac{\sigma_i}{2\epsilon_0}$

Substituting each value from the three charged sheets, we have

$E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})$

$E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

The electric field is

$E_{NET}= E_A+E_B+E_C$

$E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_{NET} = 0$

Force on each sheet is,

$F=E_{NET}\sigma ds$

$F=0$

The total force is 0

5 0

3 years ago

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

3 years ago