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Tresset [83]
2 years ago
12

Identify the right components for gsm architecture that consists of the hardware or physical equipment such as digital signal pr

ocessors, radio transceiver, display, battery, case and sim card
Engineering
1 answer:
sergiy2304 [10]2 years ago
6 0

The right components for gsm architecture that consists of the hardware or physical equipment such as digital signal processors, radio transceiver, display, battery, case and sim card is the Mobile station.

<h3>What are the 4 main components?</h3>

In GSM, a cell station includes 4 fundamental additives: Mobile termination (MT) - gives not unusualplace features consisting of: radio transmission and handover, speech encoding and decoding, blunders detection and correction, signaling and get right of entry to to the SIM. The IMEI code is connected to the MT.

Under the GSM framework, a cell tele cell smartphone is called a Mobile Station and is partitioned into  wonderful additives: the Subscriber Identity Module (SIM) and the Mobile Equipment (ME).

Read more about the mobile station:

brainly.com/question/917245

#SPJ4

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Is it possible that two types of dislocation coexist. a)-True b)- False
Svetlanka [38]

Answer:

yes it is possible

Explanation:

dislocation are if two type edge and screw dislocations

edge dislocation is a defect where an extra half plane is inside the lattice.

and screw dislocation is one in which can be assumed as the first half of the crystal slips over another.

These dislocation can coexist together where the line direction and burger vectors are  neither parallel nor perpendicular then at that condition both dislocation screw and edge will coexist

3 0
3 years ago
A tool used to drive and pull out nails​
Tju [1.3M]

Answer:

claw hammer

Explanation:

A claw hammer is a tool used for driving nails into an object or pulling nails from some other object.

6 0
2 years ago
Read 2 more answers
4. When the ESC system senses oversteer,
bagirrra123 [75]
I’m pretty sure the answer is b I’m not 100% but I think
5 0
2 years ago
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

Thus the floyds travels at 68.3^{\circ}wind speed.

7 0
4 years ago
If E = 94.2 mJ of energy is transferred when Q = 1.66 C of charge flows through a circuit element, what is the voltage across th
Anni [7]

Answer:

V = 56.8 mV

Explanation:

When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:

P = V*I

By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:

E/t = V*q/t ⇒ E = V*Q

Solving for V:

V = E/Q = 94.2 mJ /1.66 C = 56.8 mV

6 0
3 years ago
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