The position of a particle is given by s = 0.27t

What is a combination circuit? A combination circuit:

Anon25 [30]

Answer:

Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.

Example:

The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.

7 0

3 years ago

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$

$P_{real} = 2.3MW$

Real power in 3 phase

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Now the Phase Voltage is,

$V_{ph} = \frac{V}{\sqrt{3}}$

$V_{ph} = \frac{690}{\sqrt{3}}$

$V_{ph} = 398.37V$

The current phase would be,

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Rearranging,

$I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}$

Replacing,

$I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}$

$I_{ph}= 2.26kA/phase$

Therefore the current per phase is 2.26kA

6 0

3 years ago

A. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy f

NISA [10]

Answer:

a. Fraction of Atom = 2.41E-5 when T = 600K

b. Fraction of Atom = 5.03E-10 when T = 298K

Explanation:

a.

Given

T = Temperature = 600K

Qv = Energy for formation = 0.55eV/atom

To calculate the fraction of atom sites, we make use of the following formula

Nv/N = exp(-Qv/kT)

Where k = Boltzmann Constant = 8.62E-5eV/K

Nv/N = exp(-0.55/(8.62E-5 * 600))

Nv/N = 0.000024078672493307

Nv/N = 2.41E-5

b. When T = 298K

Nv/N = exp(-0.55/(8.62E-5 * 298))

Nv/N = 5.026591237904E−10

Nv/N = 5.03E-10 ----- Approximated

6 0

4 years ago

Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co

butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0

3 years ago

A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 day

erma4kov [3.2K]

D Hey will you please help me with my essay and I’ll get back to yours please ASAP

3 0

3 years ago