Is a compass a analog or a digital sensor?

(,,)=^3−^3+^3, where is the sphere ^2 + ^2 + ^2=^

miv72 [106K]

I have no clue hahahaha

7 0

3 years ago

A banked highway is designed for traffic moving at v = 88 km/h. The radius of the curve r = 314 m. show answer No Attempt 50% Pa

emmainna [20.7K]

Answer:

a) $\tan \theta = \frac{v^{2}}{g\cdot R}$

Explanation:

a) The Free Body Diagram of the vehicle and reference axis are included in the image attached below. Equations or equilibrium are presented below:

$\Sigma F_{x} = N\cdot \sin \theta = m\cdot \frac{v^{2}}{R}$

$\Sigma F_{y} = N\cdot \cos \theta - m\cdot g = 0$

Tangent of the highway can be found by dividing the first expression by the second one:

$\tan \theta = \frac{m\cdot \frac{v^{2}}{R} }{m\cdot g}$

$\tan \theta = \frac{v^{2}}{g\cdot R}$

6 0

3 years ago

Read 2 more answers

Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra

Elanso [62]

Answer: Yes

Explanation:

Surface Concentration: In electrochemistry, there is an important distinction between the concentration of a species at the electrode’s surface and its concentration at some distance from the electrode’s surface (in what we call the bulk solution). Suppose we place an electrode in a solution of Fe3+ and fix the potential at 1.00 V. At this potential Fe3+ is stable—the standard state reduction potential for Fe3+ to Fe2+ is +0.771 V, the concentration of Fe3+ remains the same at all distances from the electrode’s surface.

Bulk Concentraton: If we change the electrode’s potential to +0.500 V, the concentration of Fe3+ at the electrode’s surface decreases to approximately zero. The concentration of Fe3+ increases as we move away from the electrode’s surface until it equals the concentration of Fe3+ in bulk solution. The resulting concentration gradient causes additional Fe3+ from the bulk solution to diffuse to the electrode’s surface.

5 0

3 years ago

A compressor receives air at 290 K, 95 kPa and shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate o

aev [14]

Answer:

P2 = 3.9 MPa

Explanation:

Given that

T₁ = 290 K

P₁ = 95 KPa

Power P = 5.5 KW

mass flow rate = 0.01 kg/s

solution

with the help of table A5

here air specific heat and adiabatic exponent is

Cp = 1.004 kJ/kg K

and k = 1.4

so

work rate will be

W = m × Cp × (T2 - T1) ..........................1

here T2 = W ÷ ( m × Cp) + T1

so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290

T2 = 838 k

so final pressure will be here

P2 = P1 × $(\frac{T2}{T1})^\frac{k}{k-1}$ ..............2

P2 = 95 × $(\frac{838}{290})^\frac{1.4}{1.4-1}$

P2 = 3.9 MPa

3 0

3 years ago

The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm

oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

$m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s$