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andreyandreev [35.5K]
3 years ago
8

Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b.

DSSS and CDMA are fundamental technologies used with WiMAX BWA c. OFDM is implemented to increase spectral efficiency and to improve noise performance d. all of the statements are correct
Engineering
1 answer:
Gala2k [10]3 years ago
5 0

Answer:

d. all of the statements are correct.

Explanation:

WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.

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Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0
bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(c) 3.33 ns

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>

(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>

(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>

8 0
3 years ago
What is the significance of Saint Venant's principle?
nexus9112 [7]

Answer:

While calculating the stresses in a body since we we assume a constant distribution of stress across a cross section if the body is loaded along the centroid of the cross section , this assumption of uniformity is assumed only on the basis of Saint Venant's Principle.

Saint venant principle states that the non uniformity in the stress at the point of application of load is only significant at small distances below the load and depths greater than the width of the loaded material this non uniformity is negligible and hence a uniform stress distribution is a reasonable and correct assumption while solving the body for stresses thus greatly simplifying the analysis.

7 0
2 years ago
A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,
juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

7 0
3 years ago
Why is a metal work enclosure dangerous?
dedylja [7]

Answer:

Why is a metal work enclosure dangerous? Metalworkers are not only exposed to pollutants from metal cut ting and polishing procedures, but they are also exposed to metalworking fluids (MWF).

8 0
1 year ago
The bulk modulus of a fluid if it undergoes a 1% change in volume when subjected to a pressure change of 10,000 psi is (a) 0.01
Veseljchak [2.6K]

Answer:

The required bulk modulus is 10^{6} Psi. So, the answer is non of these.

Explanation:

Change in pressure of the fluid is directly proportional to the volumetric strain. The constant of proportionality is the bulk modulus of the fluid.

Step1

Given:

Percentage change in volume is 1%.

Change in pressure is 10000 Psi.

Calculation:

Step2

Volumetric strain is calculated as follows:

\frac{\bigtriangleup V}{V}=\frac{1}{100}

\frac{\bigtriangleup V}{V}=0.01

Step3

Bulk modulus is calculated as follows:

\frac{\bigtriangleup V}{V}=0.01

\frac{\bigtriangleup V}{V}=0.01

10000=K\times0.01

K = 1000000 Psi.

Thus, the required bulk modulus is 10^{6} Psi.

3 0
3 years ago
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