Three laws that indicate a chemical change

If a project team decided that a drugstore would earn the LT Credit—LEED for Neighborhood Development Location, what other Locat

DaniilM [7]

Answer:

None

Explanation:

It'll be impossible for the project to be eligible for any other LT credits because it can't double dip.

Double dip refers to obtaining money from two sources at the same time or by two separate accounting methods.

This is often regarded, unethical.

8 0

2 years ago

(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t

Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

$P = P_{o} + \rho gh$

Where $P$ is the absolute pressure

$P_{o}$ is the atmospheric pressure

$\rho$ is the density

$g$ is the acceleration due to gravity (Take $g = 9.8 m/s^{2}$ )

h is the height

From the question

h = 30.0 m

$\rho$ = 1.00 × 10³ kg/m³ = 1000 kg/m³

$P_{o}$ = 101.3 kPa = 101300 Pa

Using the formula

$P = P_{o} + \rho gh$

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

5 0

2 years ago

I got this information for a lab but I don't know how to do the hypothesis and the conclusion please can you guys help me with i

pochemuha

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

8 0

2 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

2 years ago

A circular coil 17.7 cm in diameter and containing 18 loops lies flat on the gound. The Earth's magnetic field at this location

iren [92.7K]

Answer:

Explanation:

Diameter of the circular coil d = 17.7 cm = 0.177m

Current in the coil, I = 7.4 A

Number of loops in the coil, N = 18

Magnetic field, B = 5.5 x 10-5 T

Angle below line, (θ) = 66°

Write the expression for the torque in the coil

Torque = N x I x B x A x sin (θ)

Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.

The cross- sectional area of the coil = (pie)(d/2)²

A = (π)(0.177/2)² = 0.0246 m²

The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north

Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°

Theta = 90 – 66 = 24°

Substitute the value to find out torque

Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m

The torque on the coil is 7.33 x 10⁻⁵ N-m

8 0

3 years ago