`int sec^(2) " z dz "`

`int sec^(2) " z dz "`
by

1 Answer

Answer :

`int sec^(2) " z dz "`
by

Related questions

Pick out the wrong statement. (A) The Reynolds analogy for mass transfer is given by Lewis relation and is applicable, when Schmidt number is one (B) Sherwood number for flow in pipes can be expressed as the ratio of the concentration gradient at wall to the overall concentration difference (C) According to film theory for equimolar counter diffusion, the mass transfer coefficient is given by DAB(B)P - 3, Q - 2 (D) The z-component of the total mass flux of a component A in binary mixture of A and B is given by - Dab'.dCA/dz

Description : Pick out the wrong statement. (A) The Reynolds analogy for mass transfer is given by Lewis relation and is applicable, when Schmidt number is one (B) Sherwood number for flow in pipes can be expressed ... flux of a component A in binary mixture of A and B is given by - Dab'.dCA/dz

Last Answer : (C) According to film theory for equimolar counter diffusion, the mass transfer coefficient is given by DAB(B)P - 3, Q - 2

`intz^(-1//3) " dz "`

Description : `intz^(-1//3) " dz "`

Last Answer : `intz^(-1//3) " dz "`

ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm

Description : ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm

Last Answer : (a) 5cm

The volume of a parallelepiped in Cartesian is a) dV = dx dy dz b) dV = dx dy c) dV = dy dz d) dV = dx dz

Description : The volume of a parallelepiped in Cartesian is a) dV = dx dy dz b) dV = dx dy c) dV = dy dz d) dV = dx dz

Last Answer : a) dV = dx dy dz

Refer to the following TAF for Zurich LSZH 261019 20018G30KT 9999 –RA SCT050 BKN080 TEMPO 23012KT 6000 –DZ BKN015 BKN030 BECMG 1518 23020G35KT 4000 RA OVC010= The lowest visibility forecast at ETA Zurich 1430 UTC is: a. 6km b. 6nm c. 4km d. 10km

Description : Refer to the following TAF for Zurich   LSZH 261019 20018G30KT 9999 –RA SCT050 BKN080 TEMPO 23012KT 6000 –DZ  BKN015 BKN030 BECMG 1518 23020G35KT 4000 RA OVC010=  The lowest visibility forecast at ETA Zurich 1430 UTC is:  a. 6km  b. 6nm  c. 4km  d. 10km

Last Answer : a. 6km

Birmingham EGBB/BHX SA0850 280850 18014KT 9999 SCT 024 BKN030 BKN045 12/08 Q1011= FC0600 280600Z 280816 190015G27KT 9999 BKN025 TEMPO 0812 5000 – DZ BKN012 BECMG 1214 19022G37= FT0400 280434Z 281212 19022G37 9999BKN025 TEMPO 1902 5000 RA BKN010 BECMG 2201 25007KT Refer to the weather information for Birmingham, above. What is the total time for which the weather is forcast?

Description : Birmingham EGBB/BHX SA0850 280850 18014KT 9999 SCT 024 BKN030 BKN045 12/08 Q1011= FC0600 280600Z 280816 190015G27KT 9999 BKN025 TEMPO 0812 5000 - DZ BKN012 BECMG 1214 19022G37= FT0400 280434Z 281212 ... information for Birmingham, above. What is the total time for which the weather is forcast?

Last Answer : 28 hours

BIRMINGHAM EGBB/BHX SA0850 280850 18014kt 9999 SCT024 BKN030 BKN045 12/08 Q1011 = FC0600 280600Z 280816 190015G27kt 9999 BKN025 TEMPO 0812 5000 – DZ BKN012 BECMG 1214 19022G37 = FT0400 280434Z 281212 19022G37 9999 BKN025 TEMPO 1902 5000 RA BKN010 BECMG 2201 25007kt Refer to weather information for Birmingham above. What is the total time for which the weather is forecast? a. 9 hours b. 18 hours c. 24 hours d. 28 hours

Description : BIRMINGHAM EGBB/BHX SA0850 280850 18014kt 9999 SCT024 BKN030 BKN045 12/08 Q1011 = FC0600 280600Z 280816 190015G27kt 9999 BKN025 TEMPO 0812 5000 - DZ BKN012 BECMG 1214 19022G37 = FT0400 280434Z 281212 19022G37 9999 ... weather is forecast? a. 9 hours b. 18 hours c. 24 hours d. 28 hours

Last Answer : d. 28 hours

Refer to the following TAF for Zurich LSZH 261019 20018G30KT 9999 –RA SCT050 BKN080 TEMPO 23012KT 6000 –DZ BKN015 BKN030 BECMG 1518 23020G35KT 4000 RA OVC010= The lowest visibility forecast at ETA Zurich 1430 UTC is: a. 6km b. 6nm c. 4km d. 10km

Description : Refer to the following TAF for Zurich LSZH 261019 20018G30KT 9999 –RA SCT050 BKN080 TEMPO 23012KT 6000 –DZ BKN015 BKN030 BECMG 1518 23020G35KT 4000 RA OVC010= The lowest visibility forecast at ETA Zurich 1430 UTC is: a. 6km b. 6nm c. 4km d. 10km

Last Answer : a. 6km

`int {1+2 tan x (tan x + sec x )}^(1//2) dx=?`

Description : `int {1+2 tan x (tan x + sec x )}^(1//2) dx=?`

Last Answer : `int {1+2 tan x (tan x + sec x )}^(1//2) dx=?` A. `log {cosec x(sec x+tan x)}+c` B. `log {sec x (sec x+ tan x)} +c` C. None of the above D.

` int sec x tan x sqrt(sec^(2) x+1)dx`

Description : ` int sec x tan x sqrt(sec^(2) x+1)dx`

Last Answer : ` int sec x tan x sqrt(sec^(2) x+1)dx`

`int sec x tan x sqrt(tan^(2) x-4) dx`

Description : `int sec x tan x sqrt(tan^(2) x-4) dx`

Last Answer : `int sec x tan x sqrt(tan^(2) x-4) dx`

`(i) int e^(x). "[log (sec x+tan x) + sec x dx "` `(ii) int (e^(-x)(cos x-sin x))/(cos^(2) x) dx`

Description : `(i) int e^(x). "[log (sec x+tan x) + sec x dx "` `(ii) int (e^(-x)(cos x-sin x))/(cos^(2) x) dx`

Last Answer : `(i) int e^(x). "[log (sec x+tan x) + sec x dx "` `(ii) int (e^(-x)(cos x-sin x))/(cos^(2) x) dx`

`int " log sin x . sec"^(2) " x dx "`

Description : `int " log sin x . sec"^(2) " x dx "`

Last Answer : `int " log sin x . sec"^(2) " x dx "`

`int " x sec"^(2) " x dx "`

Description : `int " x sec"^(2) " x dx "`

Last Answer : `int " x sec"^(2) " x dx "`

`int " x sec"^(2) " x dx "`

Description : `int " x sec"^(2) " x dx "`

Last Answer : `int " x sec"^(2) " x dx "`

`int " sec (tan"^(_1)" x ) dx "`

Description : `int " sec (tan"^(_1)" x ) dx "`

Last Answer : `int " sec (tan"^(_1)" x ) dx "`

`(i) int " x sec"^(2) " 2x dx "" "(ii) int " x sin"^(3) " x dx "`

Description : `(i) int " x sec"^(2) " 2x dx "" "(ii) int " x sin"^(3) " x dx "`

Last Answer : `(i) int " x sec"^(2) " 2x dx "" "(ii) int " x sin"^(3) " x dx "`

(i) `int(sec x* cosec x)/(log cot x) dx` (ii) `int tan^(4) x dx`

Description : (i) `int(sec x* cosec x)/(log cot x) dx` (ii) `int tan^(4) x dx`

Last Answer : (i) `int(sec x* cosec x)/(log cot x) dx` (ii) `int tan^(4) x dx`

`int(sec^(2)x)/(sqrt(tanx))dx`

Description : `int(sec^(2)x)/(sqrt(tanx))dx`

Last Answer : `int(sec^(2)x)/(sqrt(tanx))dx`

`int secx. log (sec x+ tan x ) dx `

Description : `int secx. log (sec x+ tan x ) dx `

Last Answer : `int secx. log (sec x+ tan x ) dx `

`int(secx- tan x)/(sec x+ tan x) dx`

Description : `int(secx- tan x)/(sec x+ tan x) dx`

Last Answer : `int(secx- tan x)/(sec x+ tan x) dx`

If int y = 10 then find int z = (++y * (y++ + 5));

Description : If int y = 10 then find int z = (++y * (y++ + 5));

Last Answer : Ans. Increment operator has the highest precedence. So, ++y and y++ will be evaluated starting from left. In ++y, the value of y will be incremented to 11 and then 11 will be used in the expression. When y++ is evaluated, the current ... int z = (++y * (y++ + 5)); = 11 * (11 + 5 ) = 11 * 16 = 176

How many times does the following code segment execute int x=1, y=10, z=1; do{y--; x++; y-=2; y=z; z++} while (y>1 && z<10); A) 1 B) 10 C) 5 D) infinite

Description : How many times does the following code segment execute int x=1, y=10, z=1; do{y--; x++; y-=2; y=z; z++} while (y>1 && z

Last Answer : A) 1

A three dimensional array in ‘C’ is declared as int A[x][y][z]. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer): (A) &A[0][0][0]+w(y*z*q+z*p+r) (B) &A[0][0][0]+w(y*z*p+z*q+r) (C) &A[0][0][0]+w(x*y*p+z*q+r) (D) &A[0][0][0]+w(x*y*q+z*p+r)

Description : A three dimensional array in C' is declared as int A[x][y][z]. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer): (A) &A[0][0][0]+w(y*z*q+z*p+r) (B) &A ... *q+r) (C) &A[0][0][0]+w(x*y*p+z*q+r) (D) &A[0][0][0]+w(x*y*q+z*p+r)

Last Answer : Answer: B