5. A 15-nC point charge is at the origin in free space. Calculate V₁ if point P, is located at

A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an

vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0

3 years ago

What does abbreviation vom stand for

MrMuchimi

It’s Volt Ohm Meter!

8 0

3 years ago

Read 2 more answers

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is: