What is 10 stone 11 in kg?

What is 10 stone 11 in kg?
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68.49kg
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A stone of wight 100 kg is thrown vertically upwards by a man of height y = 1.8 m with an initial velocity of 20 m/sec. The maximum height to which it will rise would be a.22.2 m b.26 m c.15.1 m d.20.2 m e.10 m

Description : A stone of wight 100 kg is thrown vertically upwards by a man of height y = 1.8 m with an initial velocity of 20 m/sec. The maximum height to which it will rise would be a.22.2 m b.26 m c.15.1 m d.20.2 m e.10 m

Last Answer : a. 22.2 m

What is 9 stone 7 pounds in kg?

Description : What is 9 stone 7 pounds in kg?

Last Answer : 7 stone 9 pounds = 48.53kg

A man lifts a stone weighting 300 g from the ground to a height of 2m and throws it horizontally with a velocity of 6 m/sec. The work done by the man is a.7.686 J b.11.286 J c.5.886 J d.8.343 J e.Tapered bearing

Description : A man lifts a stone weighting 300 g from the ground to a height of 2m and throws it horizontally with a velocity of 6 m/sec. The work done by the man is a.7.686 J b.11.286 J c.5.886 J d.8.343 J e.Tapered bearing

Last Answer : b. 11.286 J

In the book Harry Potter and the philosophers stone in chapter 11 who are the gryffindor students who play in the first quidditch match please write their name and position?

Description : In the book Harry Potter and the philosophers stone in chapter 11 who are the gryffindor students who play in the first quidditch match please write their name and position?

Last Answer : Harry Potter- SeekerKatie Bell- ChaserAlicia Spinnet- ChaserAngelina Johnson- ChaserFred Weasley- BeaterGeorge weasley- BeaterOliver Wood- Keeper

What is 11 stone pounds in pounds?

Description : What is 11 stone pounds in pounds?

Last Answer : There are 14 lb in a stone.Therefore: 14 x 11 = 154 lb

A proton beam enters magnetic field of `10^(-4) "Wb//m^(2)` normally. If the specific charge of the proton is `10^(11) C//kg` and its velocity is `10^

Description : A proton beam enters magnetic field of `10^(-4) "Wb//m^(2)` normally. If the specific charge of the proton is `10^(11) C//kg` and its velocity is `10^

Last Answer : A proton beam enters magnetic field of `10^(-4) "Wb//m^(2)` normally. If the specific charge of the proton is `10^ ... . 100 m B. 0.1 m C. 1 m D. 10 m

Momentum of electron having mass 9.1 × 10-31 kg and velocity 2.0 × 107 is A. 1.91 × 10-23 B. 2.34 × 10-23 C. 3.11 × 10-19 D. 7.88 × 10-34

Description : Momentum of electron having mass 9.1 × 10-31 kg and velocity 2.0 × 107 is A. 1.91 × 10-23 B. 2.34 × 10-23 C. 3.11 × 10-19 D. 7.88 × 10-34

Last Answer : 1.91 × 10-23

In India, which one of the following should be considered the right combination of the age in days and live body weight in kg for a lamb for its weaning ? (A) 20 days to 30 days and 6 kg to 7 kg (B) 45 days to 60 days and 8 kg to 10 kg (C) 105 days to 120 days and 18 kg to 21 kg (D) 75 days to 90 days and 12 kg to 15 kg (E) 55 days to 70 days and 9 kg to 11 kg

Description : In India, which one of the following should be considered the right combination of the age in days and live body weight in kg for a lamb for its weaning ? (A) 20 days to 30 days and 6 kg to 7 kg (B) 45 days to 60 ... (D) 75 days to 90 days and 12 kg to 15 kg (E) 55 days to 70 days and 9 kg to 11 kg

Last Answer : (C) 105 days to 120 days and 18 kg to 21 kg

A reduction of 20% in the price of sugar enables a person to get 7.8 kg more sugar for Rs.3.90. The original price of sugar per kg is a) Rs. 10 b) Rs.11.50 c) Rs. 12.5 d) Rs. 15 , e) Rs. 10.5

Description : A reduction of 20% in the price of sugar enables a person to get 7.8 kg more sugar for Rs.3.90. The original price of sugar per kg is a) Rs. 10 b) Rs.11.50 c) Rs. 12.5 d) Rs. 15 , e) Rs. 10.5

Last Answer : d) Rs. 15 ,

What is the app. Weight of 52 kg portion? (i) 10 kg (ii) 11 kg* (iii) 12 kg (iv) None of the above

Description : What is the app. Weight of 52 kg portion? (i) 10 kg (ii) 11 kg* (iii) 12 kg (iv) None of the above

Last Answer : (ii) 11 kg*

If four turms of rope around a horizontal post will hold a 100 kg weight with a pull of 1 kg, then a.107 dynes b.0.25 c.0.18 d.0.3 e.0.11

Description : If four turms of rope around a horizontal post will hold a 100 kg weight with a pull of 1 kg, then a.107 dynes b.0.25 c.0.18 d.0.3 e.0.11

Last Answer : c. 0.18

Calculate the radius of the first and second orbit of sodium atom `(Z=11)`. `(h=6.6xx10^(-34) J s, e=1.6xx10^(-19) C` and `m=9.1xx10^(-31) kg.)`

Description : Calculate the radius of the first and second orbit of sodium atom `(Z=11)`. `(h=6.6xx10^(-34) J s, e=1.6xx10^(-19) C` and `m=9.1xx10^(-31) kg.)`

Last Answer : Calculate the radius of the first and second orbit of sodium atom `(Z=11)`. `(h=6.6xx10^(-34) J s, e=1.6xx10^(-19) C` and `m=9.1xx10^(-31) kg.)`

Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6.6xx10^(-11)Nm^(2)kg^(-2)),(,"Radius of

Description : Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6.6xx10^(-11)Nm^(2)kg^(-2)),(,"Radius of

Last Answer : Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6. ... to gravity",(g),=9.8ms^(-2)):}`

Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6.6xx10^(-11)Nm^(2)kg^(-2)),(,"Radius of

Description : Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6.6xx10^(-11)Nm^(2)kg^(-2)),(,"Radius of

Last Answer : Calculate the mass and mean density of the earth from the following data : `{:(,"Gravitational constant",(G),=6. ... to gravity",(g),=9.8ms^(-2)):}`

A mass of 1 kg is attached to two identical springs each with stiffness k = 20 kN/m as shown in the figure. Under frictionless condition, the natural frequency of the system in Hz is close to * 1 point (A) 32 (B) 23 (C) 16 (D) 11

Description : A mass of 1 kg is attached to two identical springs each with stiffness k = 20 kN/m as shown in the figure. Under frictionless condition, the natural frequency of the system in Hz is close to * 1 point (A) 32 (B) 23 (C) 16 (D) 11

Last Answer : (A) 32

A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. A. 5.14 Hz B. 9.14 Hz C. 11.14 Hz D. 28.14 Hz

Description : A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. A. 5.14 Hz B. 9.14 Hz C. 11.14 Hz D. 28.14 Hz

Last Answer : C. 11.14 Hz

A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. (A) 5.14 Hz (B) 9.14 Hz (C) 11.14 Hz (D) 28.14 Hz

Description : A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. (A) 5.14 Hz (B) 9.14 Hz (C) 11.14 Hz (D) 28.14 Hz

Last Answer : (C) 11.14 Hz

One kg of carbon for theoretically complete combustion requires __________ kg of air. (A) 11.5 (B) 0.5 (C) 23 (D) 12

Description : One kg of carbon for theoretically complete combustion requires __________ kg of air. (A) 11.5 (B) 0.5 (C) 23 (D) 12

Last Answer : (A) 11.5

Dry air required to burn 1 kg of carbon completely may be around __________ kg. (A) 11 (B) 2 (C) 20 (D) 38

Description : Dry air required to burn 1 kg of carbon completely may be around __________ kg. (A) 11 (B) 2 (C) 20 (D) 38

Last Answer : (A) 11

A bed consists of particles of density 2000 kg/m3 . If the height of the bed is 1.5 metres and its porosity 0.6, the pressure drop required to fluidise the bed by air is (A) 25.61 kPa (B) 11.77 kPa (C) 14.86 kPa (D) 21.13 kPa

Description : A bed consists of particles of density 2000 kg/m3 . If the height of the bed is 1.5 metres and its porosity 0.6, the pressure drop required to fluidise the bed by air is (A) 25.61 kPa (B) 11.77 kPa (C) 14.86 kPa (D) 21.13 kPa

Last Answer : (B) 11.77 kPa

An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 metres and carries a U.D.L. of 3000 kg/m inclusive of its self weight. If the lever arm constant for the section is 0.865, the maximum intensity of shear stress, is (A) 8.3 kg/cm2 (B) 7.6 kg/cm2 (C) 21.5 kg/cm2 (D) 11.4 kg/cm2

Description : An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 metres and carries a U.D.L. of 3000 kg/m inclusive of its self weight. If the lever arm constant for the section is 0.865, the maximum intensity ... , is (A) 8.3 kg/cm2 (B) 7.6 kg/cm2 (C) 21.5 kg/cm2 (D) 11.4 kg/cm2

Last Answer : Answer: Option A

On most smooth hard surfaced roads, rolling resistance to moving vehicles, ranges from (A) 5 kg to 7 kg/tonne (B) 7 kg to 9 kg/tonne (C) 9 kg to 11 kg/tonne (D) 11 kg to 13 kg/tonne

Description : On most smooth hard surfaced roads, rolling resistance to moving vehicles, ranges from (A) 5 kg to 7 kg/tonne (B) 7 kg to 9 kg/tonne (C) 9 kg to 11 kg/tonne (D) 11 kg to 13 kg/tonne

Last Answer : Answer: Option C

One kg of carbon produces __________ kg of carbon dioxide.  A. 3/7  B. 7/3  C. 11/3  D. 3/11

Description : One kg of carbon produces __________ kg of carbon dioxide.  A. 3/7  B. 7/3  C. 11/3  D. 3/11

Last Answer : Answer: C

One kg of carbon requires 4/3 kg of oxygen and produces __________ kg of carbon monoxide gas.  A. 8/3  B. 11/3  C. 11/7  D. 7/3

Description : One kg of carbon requires 4/3 kg of oxygen and produces __________ kg of carbon monoxide gas.  A. 8/3  B. 11/3  C. 11/7  D. 7/3

Last Answer : Answer: D

One kg of carbon monoxide requires __________ kg of oxygen to produce 11/7 kg of carbon dioxide gas.  A. All  B. 4/7  C. 11/4  D. 9/7

Description : One kg of carbon monoxide requires __________ kg of oxygen to produce 11/7 kg of carbon dioxide gas.  A. All  B. 4/7  C. 11/4  D. 9/7

Last Answer : Answer: B

A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Description : A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Last Answer : 125% C.P. of sugar per kg = Rs. 4.25 C.P of the sugar per kg = [4.25×100] / 125 = Rs. 3.40. thus, the profit on 1 kg 85 paise by reducing price to 5 paise, to make same profit he must sell, 85/5 = 17kg so required ratio = 1 : 17. Answer: d)

In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Description : In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Last Answer : A)11:77:7

vIf the price of Mango is Rs. 130 per kg, Apple is Rs. 140 per kg and Orange is Rs. 80 per kg, then what is the ratio of their cost at vendor A? 1) 11 : 14 : 16 2) 39 : 28: 20 3) 39 : 28 : 21 4) 15 : 14 : 11 5) 39 : 26 : 20

Description : If the price of Mango is Rs. 130 per kg, Apple is Rs. 140 per kg and Orange is Rs. 80 per kg, then what is the ratio of their cost at vendor A? 1) 11 : 14 : 16 2) 39 : 28: 20 3) 39 : 28 : 21 4) 15 : 14 : 11 5) 39 : 26 : 20

Last Answer : 2) 39 : 28: 20

Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 0C and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge stream is mixed with superheated steam at 400 0C, 1 atm and specific volume of 3.11 m 3 /kg. Calculate the amount of superheated steam at 300 0C produced and the volumetric flow rate of the 400 0C steam.

Description : Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 0C  and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge ... amount of superheated steam at 300 0C produced and the volumetric flow rate of the  400 0C steam.

Last Answer : Solution 1. Mass balance of water 1200 + m1 = m2 ………………………………………… (1) …………………….1 mark 2. Energy balance (1200 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg)

What is the app weight of 60 kg portion ? (i) 11 kg (ii) 12 kg (iii) 13 kg* (iv) None of the above

Description : What is the app weight of 60 kg portion ? (i) 11 kg (ii) 12 kg (iii) 13 kg* (iv) None of the above

Last Answer : (iii) 13 kg*

32 kmph 11.What is the weight per meter length of 52 kg Rail? a. 54.14 kg b. 50.25 kg c. 51.89 kg* d. 53.85 kg

Description : 32 kmph 11.What is the weight per meter length of 52 kg Rail? a. 54.14 kg b. 50.25 kg c. 51.89 kg* d. 53.85 kg

Last Answer : c. 51.89 kg*

Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture? A) 8 : 5 B) 7 : 5 C) 5 : 11 D) 4 : 9 E) 5 : 7

Description : Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture? A) 8 : 5 B) 7 : 5 C) 5 : 11 D) 4 : 9 E) 5 : 7

Last Answer : Answer: E Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70=5:7

In what ratio must a person mix three kinds of dall costing him Rs 12.25,Rs 12.50 and Rs 12.75 per Kg so that the mixture may be worth Rs 12.65 per Kg? a) 11:7:1 b) 22:3:6 c) 14:3:4 d) 2:7:4 e) 3:12:32

Description : In what ratio must a person mix three kinds of dall costing him Rs 12.25,Rs 12.50 and Rs 12.75 per Kg so that the mixture may be worth Rs 12.65 per Kg? a) 11:7:1 b) 22:3:6 c) 14:3:4 d) 2:7:4 e) 3:12:32

Last Answer : E Step1: Mix dall of first and third kind to get a mixture worth Rs 12.25 per Kg. C.P of 1 Kg dall of 1st kind 1225p C.P of 1 Kg dall of 3rd kind 1275p So they must be mixed in the ... So the Mixed ratio = 3:8 Thus the quantities of dall 1st : 2nd : 3rd  = 3:12:32