How many 30 cm lengths of string can be cut from a metre of string?

How many 30 cm lengths of string can be cut from a metre of string?
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30
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Two thin lenses of focal lengths 15 cm and 30 cm resepectively are kept in contact with each other. What is the power of the combined system?

Description : Two thin lenses of focal lengths 15 cm and 30 cm resepectively are kept in contact with each other. What is the power of the combined system?

Last Answer : Two thin lenses of focal lengths 15 cm and 30 cm resepectively are kept in contact with each other. What is the power of the combined system?

A 2 kg block is dropped from a height of 4 metre onto a spring of force constant k = 1960 N/m. The spring will be compressed through a.1 cm b.4.5 cm c.3 cm d.8 cm e.107 dynes

Description : A 2 kg block is dropped from a height of 4 metre onto a spring of force constant k = 1960 N/m. The spring will be compressed through a.1 cm b.4.5 cm c.3 cm d.8 cm e.107 dynes

Last Answer : d. 8 cm

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

What fraction of a metre is 22 cm?

Description : What fraction of a metre is 22 cm?

Last Answer : It is 22/100.

A steel rod 1 metre long having square cross section is pulled under a tensile load of 8 tonnes. The extension in the rod was 1 mm only. If Esteel = 2 × 106 kg/cm2 , the side of the rod, is (A) 1 cm (B) 1.5 cm (C) 2 cm (D) 2.5 cm

Description : A steel rod 1 metre long having square cross section is pulled under a tensile load of 8 tonnes. The  extension in the rod was 1 mm only. If Esteel = 2 × 106  kg/cm2 , the side of the rod, is  (A) 1 cm  (B) 1.5 cm  (C) 2 cm  (D) 2.5 cm 

Last Answer : (C) 2 cm 

While testing a cast iron beam (2.5 cm × 2.5 cm) in section and a metre long simply supported at the ends failed when a 100 kg weight is applied at the centre. The maximum stress induced is: (A) 960 kg/cm2 (B) 980 kg/cm2 (C) 1000 kg/cm2 (D) 1200 kg/cm2 Answer: Option

Description : While testing a cast iron beam (2.5 cm × 2.5 cm) in section and a metre long simply supported at  the ends failed when a 100 kg weight is applied at the centre. The maximum stress induced is:  (A) 960 kg/cm2 (B) 980 kg/cm2 (C) 1000 kg/cm2 (D) 1200 kg/c

Last Answer : (A) 960 kg/cm

If a rectangular beam measuring 10 × 18 × 400 cm carries a uniformly distributed load such that the bending stress developed is 100 kg/cm2 . The intensity of the load per metre length, is (A) 240 kg (B) 250 kg (C) 260 kg (D) 270 kg

Description : If a rectangular beam measuring 10 × 18 × 400 cm carries a uniformly distributed load such that the bending stress developed is 100 kg/cm2 . The intensity of the load per metre length, is (A) 240 kg (B) 250 kg (C) 260 kg (D) 270 kg

Last Answer : (B) 250 kg

A 8 metre long simply supported rectangular beam which carries a distributed load 45 kg/m. experiences a maximum fibre stress 160 kg/cm2 . If the moment of inertia of the beam is 640 cm4 , the overall depth of the beam is (A) 10 cm (B) 12 cm (C) 15 cm (D) 18 cm

Description : A 8 metre long simply supported rectangular beam which carries a distributed load 45 kg/m. experiences a maximum fibre stress 160 kg/cm2 . If the moment of inertia of the beam is 640 cm4 , the overall depth of the beam is (A) 10 cm (B) 12 cm (C) 15 cm (D) 18 cm

Last Answer : (A) 10 cm

Area of steel required per metre width of pavement for a length of 20 m for design wheel load 6300 kg and permissible stress in steel 1400 kg/cm2 , is (A) 70 kg/sq cm (B) 80 kg/sq cm (C) 90 kg/sq cm (D) 100 kg/sq cm

Description : Area of steel required per metre width of pavement for a length of 20 m for design wheel load 6300 kg and permissible stress in steel 1400 kg/cm2 , is (A) 70 kg/sq cm (B) 80 kg/sq cm (C) 90 kg/sq cm (D) 100 kg/sq cm

Last Answer : Answer: Option C

The taper of precast concrete pile should not be more than (A) 1 cm per metre length (B) 2 cm per metre length (C) 4 cm per metre length (D) 5 cm per metre length

Description : The taper of precast concrete pile should not be more than (A) 1 cm per metre length (B) 2 cm per metre length (C) 4 cm per metre length (D) 5 cm per metre length

Last Answer : Answer: Option B

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : We know that, each angle of a rectangle is right angle (i.e., 90°) and its opposite sides are equal and parallel. To construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm, use the ... AD and CD. Thus, ABCD is the required rectangle with adjacent sides of length 5 cm and 3.5 cm.

A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Description : A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Last Answer : we know that , all sides of a rhombus are equal and the diagonals of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4cm and 6cm ... ) From Eqs. (i) and (ii), AB = BC = CD = DA Hence , ABCD is a rhombus.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Description : A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Last Answer : We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps. 1.Draw ... B and D. 6.Now, join AB, BC, CD, and DA . Thus, ABCD is the required rhombus.

Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- Yes, because in each case sum of two sides is greater than the third side.

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Description : The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Last Answer : There are two parallel chords of length 8 cm and 6 cm. The distance between center and shortest chord (6cm chord) is 4cm. So, the perpendicular distance from the center to the shortest chord is 4cm. The ... 32+42 =5. Since the radius is 5. The distance from center to largest chord is 52−42 =3.

The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Last Answer : Let, a = 7 cm, b = 13 cm, c = 12 cm ∴ s = (a + b + c)/2 = (7 +13 +12)/2 = 32/2 = 16 cm Area of △ ABC = under root( √s(s -a) (s - b)(s -c)) = under root( √16(16 - 7)(16 - 13)(16 - 12) = ... 24 √3 cm2 Also, Area of △ ABC = 1/2AC.BD 24 √3 = 1/2 x 12 x BD ⇒ BD = (24 √3 x 2)/12 = 4 √3 cm

The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Description : The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Last Answer : For △BCD: Let a = 17 cm, b = 12 cm, c = 25 cm So its semi-perimeter, s = (a + b + c)/2 = (17 + 12 + 25)/2 = 27 cm ∴ Area of △BCD = root under(√(s -a)(s - b)(s - c)) = ... △BCD = 2 x 90 = 180 cm2 Also, area of parallelogram ABCD = DC x AE ∴ 180 = 12 x AE ⇒ AE = 180/12 = 15 cm

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Description : The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Last Answer : (b) 72 cm2Here sm = \(rac{9+12+15}{2}\) = 18 cm, where lengths of medians are m1 = 9 cm, m2 = 12 cm, m3 = 15 cm.∴ Area of triangle = \(rac{4}{3}\sqrt{18(18-9)(18-12)(18-15)}\) cm2= \(rac{4}{3}\sqrt{18 imes9 imes6 imes3}\) cm2 = \(rac{4}{3}\) x 9 x 6 cm2 = 72 cm2.

What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Description : What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Last Answer : (d) 6 cmLet a = 35 cm, b = 44 cm, c = 75 cm. Thens = \(rac{a+b+c}{2}\) = \(rac{34+44+75}{2}\) cm = 77 cm∴ Area if triangle = \(\sqrt{77(77-35)(77-44)(77-75)}\) cm2= \(\sqrt{77 imes42 ... ) cm2 = 462 cm2∴ Radius of incircle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \(rac{462}{77}\) cm = 6 cm.

If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel -Maths 9th

Description : The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel -Maths 9th

Last Answer : Area of trapezium =12×h(AB+CD) =12×8×(8+14)=12×8×(8+14) =4×22=88cm2=4×22=88cm2 = Volume of prism = Height of prism ×× area of base ⇒height×88=1056 (given)⇒height×88=1056 (given) ⇒height×88=105688⇒height×88=105688 ⇒12cm =12×h(AB+CD)

A soccer ball is packaged in a cubic box with side lengths of 22 cm. Assuming that the ball touches all faces of the box, calculate what percentage of the volume of the box is wasted space Write it step by step pls?

Description : A soccer ball is packaged in a cubic box with side lengths of 22 cm. Assuming that the ball touches all faces of the box, calculate what percentage of the volume of the box is wasted space Write it step by step pls?

Last Answer : one centimetre

A soccer ball is packaged in a cubic box with side lengths of 22 cm. Assuming that the ball touches all faces of the box, calculate what percentage of the volume of the box is wasted space?

Description : A soccer ball is packaged in a cubic box with side lengths of 22 cm. Assuming that the ball touches all faces of the box, calculate what percentage of the volume of the box is wasted space?

Last Answer : one centimetre

How many 10 cm lengths are there in 5 m?

Description : How many 10 cm lengths are there in 5 m?

Last Answer : There are 50.

ails are available in various lengths ranging from a ) 1.00 to 15 cm b) 1.25 to 15 cm c) 1.5 to 15 cm d) 2.5 to 15 cm

Description : ails are available in various lengths ranging from a ) 1.00 to 15 cm b) 1.25 to 15 cm c) 1.5 to 15 cm d) 2.5 to 15 cm

Last Answer : b) 1.25 to 15 cm

The difference in the lengths of an arc and its subtended chord on the earth surface for a distance of 18.2 km, is only (A) 1 cm (B) 5 cm (C) 10 cm (D) 100 cm

Description : The difference in the lengths of an arc and its subtended chord on the earth surface for a distance of 18.2 km, is only (A) 1 cm (B) 5 cm (C) 10 cm (D) 100 cm

Last Answer : (C) 10 cm

In case of Raymond pile (A) Lengths vary from 6 m to 12 m (B) Diameter of top of piles varies from 40 cm to 60 cm (C) Diameter of pile at bottom varies from 20 cm to 28 cm (D) All the above

Description : In case of Raymond pile (A) Lengths vary from 6 m to 12 m (B) Diameter of top of piles varies from 40 cm to 60 cm (C) Diameter of pile at bottom varies from 20 cm to 28 cm (D) All the above

Last Answer : Answer: Option D

A wire whose resistance is 90 ohm is cut into 3 pieces of equal lengths which are then arranged in parallel. Calculate the resistance of the combinati

Description : A wire whose resistance is 90 ohm is cut into 3 pieces of equal lengths which are then arranged in parallel. Calculate the resistance of the combinati

Last Answer : A wire whose resistance is 90 ohm is cut into 3 pieces of equal lengths which are then ... in parallel. Calculate the resistance of the combination.

BMR of healthy adult men is about (A) 30 kcal/hour/square metre (B) 35 kcal/hour/square metre (C) 40 kcal/hour/square metre (D) 45 kcal/hour/square metre

Description : BMR of healthy adult men is about (A) 30 kcal/hour/square metre (B) 35 kcal/hour/square metre (C) 40 kcal/hour/square metre (D) 45 kcal/hour/square metre

Last Answer : Answer : C

In precision theodolite traverse if included angles are read twice and the mean reading accepted using both verniers having a least count of 30". Assuming the instrument to be in perfect adjustment, linear measurements correct to 6 mm per 30 metre tape duly corrected for temperature, slope and sag, the angular error of closure not to exceed (where n is the number of traverse legs) (A) 50" n(B) 30" n(C) 60" n (D) None of these

Description : In precision theodolite traverse if included angles are read twice and the mean reading accepted using both verniers having a least count of 30". Assuming the instrument to be in perfect adjustment, linear measurements correct to 6 mm per ... legs) (A) 50" n (B) 30" n (C) 60" n (D) None of these

Last Answer : (A) 50" n

A shaft 9 m long is subjected to a torque 30 t-m at a point 3 m distant from either end. The reactive torque at the nearer end will be (A) 5 tonnes metre (B) 10 tonnes metre (C) 15 tonnes metre (D) 20 tonnes metre

Description : A shaft 9 m long is subjected to a torque 30 t-m at a point 3 m distant from either end. The reactive torque at the nearer end will be (A) 5 tonnes metre (B) 10 tonnes metre (C) 15 tonnes metre (D) 20 tonnes metre

Last Answer : (D) 20 tonnes metre

A pile of length carrying a uniformly distributed load per metre length is suspended at the centre and from other two points 0.15 L from either end ; the maximum hogging moment will be (A) WL²/15 (B) WL²/30 (C) WL²/60 (D) WL²/90

Description : A pile of length carrying a uniformly distributed load per metre length is suspended at the centre and from other two points 0.15 L from either end ; the maximum hogging moment will be (A) WL²/15 (B) WL²/30 (C) WL²/60 (D) WL²/90

Last Answer : Answer: Option D

Pick up the correct statement from the following: (A) Long tangent sections exceeding 3 km in length should be avoided (B) Curve length should be at least 150 metres for a deflection angle of 5 degree (C) For every degree decrease in the deflection angle, 30 metre length of curve to be increased (D) All the above

Description : Pick up the correct statement from the following: (A) Long tangent sections exceeding 3 km in length should be avoided (B) Curve length should be at least 150 metres for a deflection angle of 5 ... decrease in the deflection angle, 30 metre length of curve to be increased (D) All the above

Last Answer : Answer: Option D

The electrostatic force of repulsion between two electrons at 1 metre is: (a) 9 × 109 N (b) 1.44 × 10−9 N (c) 2.30 × 10−28 N (d) 1 N

Description : The electrostatic force of repulsion between two electrons at 1 metre is: (a) 9 × 109 N (b) 1.44 × 10−9 N (c) 2.30 × 10−28 N (d) 1 N

Last Answer : (c) 2.30 × 10−28 N

Expansion Joints in masonry walls are provided in wall lengths usater than (A) 10 m (B) 20 m (C) 30 m (D) 40 m

Description : Expansion Joints in masonry walls are provided in wall lengths usater than (A) 10 m (B) 20 m (C) 30 m (D) 40 m

Last Answer : Answer: Option D

Two trains A and B are moving in opposite directions at 180 km/hr and 270 km/hr. their lengths are 3.30 km and 2.7 km respectively. The time taken by the slower train to cross the faster train in second is. A) 48sec B) 13 sec C) 38sec D) 24sec

Description : Two trains A and B are moving in opposite directions at 180 km/hr and 270 km/hr. their lengths are 3.30 km and 2.7 km respectively. The time taken by the slower train to cross the faster train in second is. A) 48sec B) 13 sec C) 38sec D) 24sec

Last Answer : ANSWER: A Explanation: Speed of 1st train = 180 km/hr  Speed of 2nd train = 270 km/hr  Relative speed = (180 + 270)km/hr  = 450 km/hr  = (450 * 5 / 180)m/s  = (2250 / 18)m/s  =125 m/s  Distance covered = (3.30 + 2.70)km  = 6km = 6000m  Required time = 6000 / 125  = 48sec

Two goods trains having equal lengths, take 30 seconds and 45 seconds respectively to cross a telegraph post. If the length of each train is 360 meters, in what time (in seconds) will they cross each other when traveling in opposite direction? A) 36 B) 12 C) 24 D) 48

Description :  Two goods trains having equal lengths, take 30 seconds and 45 seconds respectively to cross a telegraph post. If the length of each train is 360 meters, in what time (in seconds) will they cross each other when traveling in opposite direction? A) 36 B) 12 C) 24 D) 48

Last Answer : ANSWER:A  Explanation:  Speed of train = distance /time  = 360/30 m/s  = 12 m/s  Speed of train 2 = distance /time  = 360 /45 m/s  = 8 m/s  If 2 trains are moving in opposite ... +360 = 720 m  Time = distance/speed  = (720/20) = 36 Seconds Hence the required answer is 36 seconds

A stone weighing 20 N is whirling in a vertical circle at the extremity of a string 100 cm long. The tension in the string will be zero a.at the highest position b.at the lowest position c.at the midway position d.at none of the above e.107 dynes

Description : A stone weighing 20 N is whirling in a vertical circle at the extremity of a string 100 cm long. The tension in the string will be zero a.at the highest position b.at the lowest position c.at the midway position d.at none of the above e.107 dynes

Last Answer : a. at the highest position

A piece of string is 2 1/2 m long.If John cuts off one piece 60 cm long and another 80 cm long, what is the length of the remaining piece of string?

Description : A piece of string is 2 1/2 m long.If John cuts off one piece 60 cm long and another 80 cm long, what is the length of the remaining piece of string?

Last Answer : 70 CM

To prevent movement of moisture from sub-grade to road pavement at the same level as that of water-table, thickness of a cut off layer of coarse sand, is (A) 15 cm (B) 20 cm (C) 30 cm (D) 45 cm

Description : To prevent movement of moisture from sub-grade to road pavement at the same level as that of water-table, thickness of a cut off layer of coarse sand, is (A) 15 cm (B) 20 cm (C) 30 cm (D) 45 cm

Last Answer : Answer: Option A

To prevent movement of moisture from sub-grade to road pavement on embankments about 0.6 m to 1 m higher than water table, the thickness of cut off layer of coarse sand used, is (A) 15 cm (B) 20 cm (C) 30 cm (D) None of these

Description : To prevent movement of moisture from sub-grade to road pavement on embankments about 0.6 m to 1 m higher than water table, the thickness of cut off layer of coarse sand used, is (A) 15 cm (B) 20 cm (C) 30 cm (D) None of these

Last Answer : Answer: Option D

Cut a string into pieces (PHP! ;-) )

Description : Cut a string into pieces (PHP! ;-) )

Last Answer : You could use substr while looping up to the length of the string and using array_push to build your $string_array.