Description : In the formula `S=sqrt((p^(2)+q)/(p-q))` make q the subject.
Last Answer : In the formula `S=sqrt((p^(2)+q)/(p-q))` make q the subject.
Description : In the formula, `x=(p+5q)/(sqrt(3)+2q)`, make q the subject.
Last Answer : In the formula, `x=(p+5q)/(sqrt(3)+2q)`, make q the subject.
Description : In the formula `k=((a-b)/(a+b))^(2)`, find the value of b if `k=36` and a=3, making b the subject.
Last Answer : In the formula `k=((a-b)/(a+b))^(2)`, find the value of b if `k=36` and a=3, making b the subject.
Description : The diagonal of a rectangle is 10 units. The formula to find the diagonal is `d=sqrt(l^(2)+b^(2))`. Where l and b are length and breadth respectively.
Last Answer : The diagonal of a rectangle is 10 units. The formula to find the diagonal is `d=sqrt(l^(2)+b ... breadth respectively. (10+b)(10-b) is equal to______.
Description : In the formula `T=2pisqrt((l)/(g)),T` is ____proportional to `sqrt(l)`.
Last Answer : In the formula `T=2pisqrt((l)/(g)),T` is ____proportional to `sqrt(l)`.
Description : `" if " int (sin 2x- cos 2x) dx=(1)/(sqrt(2)) sin (2x-k)+c " then " k=?`
Last Answer : `" if " int (sin 2x- cos 2x) dx=(1)/(sqrt(2)) sin (2x-k)+c " then " k=?` A. `-(5pi)/(4)` B. `(pi)/(4)` C. `-(pi)/(4)` D.
Description : A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s. a. 11,120 N b. 11,320 N c. 11,420 N d. 11520 N formula: F= ma / k
Last Answer : 11,120N
Description : Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg. a. 20 kg b. 19 kg c. 18 kg d. 17 kg formula: m=Fg k / g
Last Answer : 18 kg
Description : A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k ) a. 48.75 N b. 50 N c. 45 N d. None of the above
Last Answer : 48.75 N
Description : A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k ) a. 543680 N/m2 b. 543.68 kPa (gauge) c. Both a & b d. None of the above
Last Answer : Both a & b
Description : Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane. A. 2474.23 N B. 2574.23 N C. 3474.23 N D. 2374.23 N Formula : M = wk / g F = ma /k
Last Answer : 2474.23 N
Description : A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2 A. 4474.23 N B.5484.23 N C.4495.23 N D.5488.23 N Formula: F = m/k (a +g)
Last Answer : 4474.23 N
Description : `int((a+bsin^(-1)x)^(n))/(sqrt(1-x^(2)))dx`
Last Answer : `int((a+bsin^(-1)x)^(n))/(sqrt(1-x^(2)))dx`
Description : Let `alphaa n dbeta` be nonzero real numbers such that `2(cosbeta-cosalpha)+cosalphacosbeta=1` . Then which of the following is/are true? `sqrt(3)tan(
Last Answer : Let `alphaa n dbeta` be nonzero real numbers such that `2(cosbeta-cosalpha)+cosalphacosbeta=1` . Then which of the ... (alpha)/(2))+tan((beta)/(2))=0`
Description : If `(a)/(b)=(b)/(c)`, then make c as the subject of the formula.
Last Answer : If `(a)/(b)=(b)/(c)`, then make c as the subject of the formula.
Description : In `v^(2)-u^(2)=2as`, make a as the subject of the formula.
Last Answer : In `v^(2)-u^(2)=2as`, make a as the subject of the formula.
Description : A wire of length a units is cut into two equal pieces and make a circle of radius r with one piece. Frame the formula by making r the subject.
Last Answer : A wire of length a units is cut into two equal pieces and make a circle of radius r with one piece. Frame the formula by making r the subject.
Description : Make b the subject for the formula `f=(a+b)/(a)`.
Last Answer : Make b the subject for the formula `f=(a+b)/(a)`.
Description : Fahrenheit temperature F is 32 more tha ninefifths of the centrigrade temperature C. Frame the formula making F as the subject.
Last Answer : Fahrenheit temperature F is 32 more tha ninefifths of the centrigrade temperature C. Frame the formula making F as the ... C-32` D. `F=32xx(9)/(5)C`
Description : A is the area of the square with side S units. Write the formula of area making S as the subject.
Last Answer : A is the area of the square with side S units. Write the formula of area making S as the subject.
Description : A is the area of the rectangle with length l units and breadth b units. Rewrite the formula for area making l as the subject.
Last Answer : A is the area of the rectangle with length l units and breadth b units. Rewrite the formula for area making l as the subject.
Description : A is the area of a circle with radius r units. Write the formula of area, making r as the subject.
Last Answer : A is the area of a circle with radius r units. Write the formula of area, making r as the subject.
Description : The formula with subject C is `C=(5F-160)/(9)`. Then what is the formula with subject F?
Last Answer : The formula with subject C is `C=(5F-160)/(9)`. Then what is the formula with subject F?
Description : Evaluate: `lim_(x rarr 0) (sqrt(5+x)-sqrt(5-x))/(sqrt(10+x)-sqrt(10-x))`.
Last Answer : Evaluate: `lim_(x rarr 0) (sqrt(5+x)-sqrt(5-x))/(sqrt(10+x)-sqrt(10-x))`.
Description : If `[root(9)((2/3)^5)]^(sqrt(x-5))` = `a^0`, find the value of `x`.
Last Answer : If `[root(9)((2/3)^5)]^(sqrt(x-5))` = `a^0`, find the value of `x`.
Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3
Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3
Description : In an `LC` circuit shows in Fig. `C = 1 F`, `L = 4 H`. At time `t = 0`, charge in the capacitor is `4 C` and it is decreasing at the rate of `sqrt(5)
Last Answer : In an `LC` circuit shows in Fig. `C = 1 F`, `L = 4 H`. At time `t = 0`, charge in the capacitor is `4 ... ))` C. `2tan^(-1)((2)/(3)) D. None of these
Last Answer : In an `LC` circuit shows in Fig. `C = 1 F`, `L = 4 H`. At time `t = 0`, charge in the capacitor is `4 ... statement. A. 6 C B. 8 C C. 10 C D. 12 C
Description : The drainage area of a town is 12 hectares. Its 40% area is hard pavement ( K = 0.85), the 40% area is unpaved streets (K = 0.20) and the remaining is wooded areas (K = 0.15). Assuming the time of concentration ... 60) the maximum run off is A. 0.10 cumec B. 0.12 cumec C. 0.15 cumec D. 0.20 cumec
Last Answer : ANS: C
Description : For calculating the evaporation rate over a reservoir surface E = 0.771 (1.465 - 0.00732B) (0.44 - 0.007375 V) (pe - pa), the equation is given by (A) Roohwer's, formula in M.K.S. (B) Roohwer's formula in F.P.S. (C) Dalton's formula in F.P.S. (D) Dalton's formula in M.K.S
Last Answer : Answer: Option A
Description : For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k? a. 1.36 b. 1.37 c. 1.38 d. 1.39 formula: k= R / cv+1
Last Answer : 1.38
Description : In the above problem, find the potential energy of the mass with respect to datum. (Formula: P = mgz/k ) a. 4875 j b. 0.51 j c. 0.46 j d. None of the above
Last Answer : 4875 j
Description : Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water? A. 0.043 K/s B. 0.179 K/s C. 0.357 K/s D. 1.50 K/s Formula: Q = mcv ( T)
Last Answer : 0.179 K/s
Description : `int_(0)^(pi//4) sqrt(cot x dx) =?`
Last Answer : `int_(0)^(pi//4) sqrt(cot x dx) =?` A. `(Pi sqrt(2))/(4)+(1)/(sqrt(2)) log (sqrt(2)-1)` B. `(- ... (pisqrt(2))/(4) -(1)/(sqrt(2))log (sqrt(2)-1)` D.
Description : `int_(0)^(1//2) (x sin^(-1)x)/(sqrt(1-x^(2)))dx=?`
Last Answer : `int_(0)^(1//2) (x sin^(-1)x)/(sqrt(1-x^(2)))dx=?` A. `(pi sqrt(3))/(12)+(1)/(2)` B. `(pisqrt(3))/(12)-(1)/(2)` C. `(pisqrt(3))/(12) -(1)/(2)` D.
Description : `int_(0)^(1) (x)/(sqrt(1+x^(2)))dx=?`
Last Answer : `int_(0)^(1) (x)/(sqrt(1+x^(2)))dx=?` A. `sqrt(2)-1` B. `sqrt(2)` C. `-sqrt(2)` D.
Description : `int_(0)^(1) sqrt((1-x)/(1+x)) dx=?`
Last Answer : `int_(0)^(1) sqrt((1-x)/(1+x)) dx=?` A. `(pi)/(2)+1` B. `(pi)/(2) -1` C. None of these D.
Description : `int_(0)^(1) x sqrt((1-x^(2))/(1+x^(2)))dx`
Last Answer : `int_(0)^(1) x sqrt((1-x^(2))/(1+x^(2)))dx`
Description : `int_(0)^(2) sqrt((2+x)/(2-x)) dx`
Last Answer : `int_(0)^(2) sqrt((2+x)/(2-x)) dx`
Description : `int_(0)^(1//sqrt(2)) (sin^(-1))/((1-x^(2))^(3//2))dx`
Last Answer : `int_(0)^(1//sqrt(2)) (sin^(-1))/((1-x^(2))^(3//2))dx`
Description : `int_(0)^(1) (x sin^(-1)x)/(sqrt(1+2x)^(2))dx`
Last Answer : `int_(0)^(1) (x sin^(-1)x)/(sqrt(1+2x)^(2))dx`
Description : `int_(0)^(a) (x)/(sqrt(a^(2)-x^(2)))dx`
Last Answer : `int_(0)^(a) (x)/(sqrt(a^(2)-x^(2)))dx`
Description : `int_(0)^(pi//2) sqrt(1- cos 2x) dx`
Last Answer : `int_(0)^(pi//2) sqrt(1- cos 2x) dx`
Description : `int_(0)^(1) (1)/(sqrt(1-x^(2)))dx`
Last Answer : `int_(0)^(1) (1)/(sqrt(1-x^(2)))dx`
Description : `int_(0)^(pi//6) sqrt(1-sin 2x) dx`
Last Answer : `int_(0)^(pi//6) sqrt(1-sin 2x) dx`
Description : Evaluate: `lim_underset(x rarr 0) (sqrt(1+x+x^(2)+x^(3))-1)/(x)`.
Last Answer : Evaluate: `lim_underset(x rarr 0) (sqrt(1+x+x^(2)+x^(3))-1)/(x)`.
Description : Using differentials, find the approximate values of the following: `sqrt(0.24)`
Last Answer : Using differentials, find the approximate values of the following: `sqrt(0.24)`
Description : Using differentials, find the approximate value of `sqrt(0. 48)`
Last Answer : Using differentials, find the approximate value of `sqrt(0. 48)`
Description : Let `S={x in R: x ge 0 and 2|(sqrt(x)-3|+sqrt(x)(sqrt(x)-6)+6=0}` then S (1) is an empty set (2) contains exactly one element (3) contains exact;y two
Last Answer : Let `S={x in R: x ge 0 and 2|(sqrt(x)-3|+sqrt(x)(sqrt(x)-6)+6=0}` then S ... four elements C. is an empty set. D. contains exactly one elements