Where is the highest value of gravitational acceleration 'g ' ?

Where is the highest value of gravitational acceleration 'g ' ?
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Answer :

The value of gravitational acceleration 'g' is highest in the polar region
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Related questions

What is the value of gravitational acceleration ( g) in the center of the earth ?

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Last Answer : The value of gravitational acceleration ( g) in the center of the earth is zero (0).

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Description : What is the value of gravitational acceleration g ?

Last Answer : Gravitational acceleration g has a value of 9.8 m / s.

The average person passes out at an acceleration of 7g (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time (in seconds) would be required for the car to accelerate from rest to 56.3 miles per hour?

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Last Answer : I think the Fluther policy is not to help students with their homework, but I always hate this kind of problem and it seems to me that the g is just in there to provide some background and to confuse you.

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The value of G, the universal gravitational constant, was measured experimentally by: w) Newton x) Cavendish y) Copernicus z) Kepler

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What is meant by gravitational acceleration value of 9.8ms 2 ?

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As we go from Equator to North pole the value of ‘g’, the acceleration due to gravity (1) remains the same (2) decreases (3) increases (4) None of the above

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Description : What is the unit of gravitational constant G ?

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Dimensions of gravitational constant ‘G’ are: A. [M]-1 [L]3 [T]-2 B. [M] [L]3 [T]-2 C. [M]-1 [L]2 [T]-1 D. [M] [L]-1 [T]2

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Last Answer : [M]-1 [L]3 [T]-2

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The value of the universal gravitational constant, G is (a) 6.67 × 1011 Nm2/kg2 (b) 6.67 × 10+9 Nm2/kg2 (c) 6.67 × 10–9 Nm2/kg2 (d) 6.67 × 10–11 Nm2/kg2

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Gravitational constant (G) is an example of (a) Non-dimensional constant (b) Dimensional constant (c) Numeric constant (d) Quantity without dimensions

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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

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

At the top of the first hill the cart has the highest (Gravitational) Potential Energy?

Description : At the top of the first hill the cart has the highest (Gravitational) Potential Energy?

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A weight of 4.9 N rests on a rough horizontal floor that has a coefficient of friction of 0.5. If g = 9.8 m/s2, the smallest horizontal force that gives the weight an acceleration of 4m/s2 is most nearly a.2.5 N b.3 N c.2 N d.4.5 N e.4.9 N

Description : A weight of 4.9 N rests on a rough horizontal floor that has a coefficient of friction of 0.5. If g = 9.8 m/s2, the smallest horizontal force that gives the weight an acceleration of 4m/s2 is most nearly a.2.5 N b.3 N c.2 N d.4.5 N e.4.9 N

Last Answer : d. 4.5 N

Show that acceleration due to gravity at height h above the Earth’s surface is gh = g(R/R + h)^2

Description : Show that acceleration due to gravity at height h above the Earth’s surface is gh = g(R/R + h)^2

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Acceleration due to gravity (g) is mini- mum at?

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Acceleration due to gravity (g) is maxi- mum at?

Description : Acceleration due to gravity (g) is maxi- mum at?

Last Answer : polar regions

At the Center of earth Acceleration due to gravity (g) is?

Description : At the Center of earth Acceleration due to gravity (g) is?

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Critical speed of rotation, N (in rps - rotation per second) of a trammel is equal to (where, g = acceleration due to gravity = 9.81 m/sec 2 and, r = radius of trammel, metre.) (A) (1/2π). √(g/r) (B) (1/π). √(g/r) (C) ½ √(g/r) (D) 2π. √(g/r)

Description : Critical speed of rotation, N (in rps - rotation per second) of a trammel is equal to (where, g = acceleration due to gravity = 9.81 m/sec 2 and, r = radius of trammel, metre.) (A) (1/2π). √(g/r) (B) (1/π). √(g/r) (C) ½ √(g/r) (D) 2π. √(g/r)

Last Answer : (A) (1/2π). √(g/r)

The static deflection of a spring under gravity, when a mass of 1 kg is suspended from it, is 1 mm. Assume the acceleration due to gravity g = 10 m/s^2. The natural frequency of this spring-mass system (in rad/s) is A 100 B 150 C 200 D 250

Description : The static deflection of a spring under gravity, when a mass of 1 kg is suspended from it, is 1 mm. Assume the acceleration due to gravity g = 10 m/s^2. The natural frequency of this spring-mass system (in rad/s) is A 100 B 150 C 200 D 250

Last Answer : A 100

A lift of weight W is lifted by a rope with an acceleration f. If the area of cross-section of the rope is A, the stress in the rope is (A) [W (1 + f/G)]/A (B) (1 - g/f)/A (C) [W (2 + f/G)]/A (D) [W (2 + g/f)]/A

Description : A lift of weight W is lifted by a rope with an acceleration f. If the area of cross-section of the rope  is A, the stress in the rope is  (A) [W (1 + f/G)]/A (B) (1 - g/f)/A (C) [W (2 + f/G)]/A (D) [W (2 + g/f)]/A

Last Answer : (A) [W (1 + f/G)]/A

If is the speed of a locomotive in km per hour, g is the acceleration due to gravity, is the distance between running faces of the rails and is the radius of the circular curve, the required super elevation is (A) gV²/GR (B) Rg/GV² (C) GR/gV² (D) GV²/gR

Description : If is the speed of a locomotive in km per hour, g is the acceleration due to gravity, is the distance between running faces of the rails and is the radius of the circular curve, the required super elevation is (A) gV²/GR (B) Rg/GV² (C) GR/gV² (D) GV²/gR

Last Answer : (D) GV²/gR

If V is speed of a moving vehicle, r is radius of the curve, g is the acceleration due to gravity, W is the width of the carriageway, the super elevation is (A) WV/gr (B) W²V/gr (C) WV²/gr (D) WV/gr²

Description : If V is speed of a moving vehicle, r is radius of the curve, g is the acceleration due to gravity, W is the width of the carriageway, the super elevation is (A) WV/gr (B) W²V/gr (C) WV²/gr (D) WV/gr²

Last Answer : Answer: Option C

Would the gravitational force of a paper plate the size of our solar system crush someone standing at the middle of the plate?

Description : Would the gravitational force of a paper plate the size of our solar system crush someone standing at the middle of the plate?

Last Answer : answer:Well, calculate it by integration. Say, a paper plate weighs 10 gram per square meter. Do not forget that there is symmetry, so the force components parallel with the plate cancel. Calculate your integral.

Which scientist gave the formula of gravitational ball ?

Description : Which scientist gave the formula of gravitational ball ?

Last Answer : Scientist Newton gives the formula for the gravitational ball.

Explain that gravitational force is an untouchable force ?

Description : Explain that gravitational force is an untouchable force ?

Last Answer : The force that acts without direct contact of two objects is called untouchable force. The force of attraction between any two objects in the universe is called gravity. Touch is not required for ... solar system revolve around the sun without touching it. So the force of gravity is untouchable.