Weight = mass * acceleration due to gravity.
In the SI system, g = 9.8 m/s^2.
Hence, your weight at the equator will be 9.8xm Newton. ( I dont know your mass. you see. so I'm writing 'm')
Cheers!
it would weigh the same because the mortor comes from earth
no
This would be known as the Equator
If you assume thatequatriol = equatorial raduis = radius then the equatorial radius is the radius of the spheroid measured at its equator. It would be = length of the equator/(2*pi)
The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.
Mass would be the same at the equator and at the pole - except for an insignificant change due to the General Theory of Relativity. Weight would be more at the pole.
To calculate weight at the equator and pole, you need to consider the effect of gravity. At the equator, the centrifugal force due to the Earth's rotation slightly reduces weight compared to the pole. The weight at the pole is higher because the centrifugal force is lower there. However, the difference in weight between the pole and equator is very small and often negligible for everyday purposes.
In that case, your weight remains absolutely constant and does not budge one iota.
The weight of an object is less at the equator compared to the poles due to the centripetal force produced by the Earth's rotation. At the equator, this force partially counteracts the force of gravity, effectively reducing the object's weight. This difference in weight is more noticeable for objects with larger mass.
The weight of an object changes when it is moved from the equator to the poles due to the variation in gravitational force caused by the Earth's rotation. The force of gravity is slightly stronger at the poles compared to the equator, leading to a small change in weight.
The weight of an object would not change when taken from Delhi to the pole. Weight is the force of gravity acting on an object, and this force remains constant regardless of location. The object's mass would remain the same, but its weight may be perceived differently due to variations in gravity strength at different locations.
That would depend on exactly where on the the equator you are. If you are on the equator by Brazil then you would have to steer south. If you are on the equator nea PNG the you have to steer east.
That would depend on exactly where on the the equator you are. If you are on the equator by Brazil then you would have to steer south. If you are on the equator nea PNG the you have to steer east.
Yes. The mass of an object will stay the same, regardless of the gravity that is effecting it. But the weight of an object depends on the apparent gravity. At the poles you would weigh more than at the equator due to the earths spin. At the equator you might weigh up to 0.3% less than atthe poles. Other factors effect the local gravity such as the density of the rock beneath the person, more dense rock will give a higher gravitational field. The height above the surface will also reduce the apparent gravity.
At the poles you are closer (slightly) to the centre of gravity of the earth, than at the equator. So a spring-balance weighing machine would register a slightly higher weight at the pole. Notice that weights for a see-saw balance are similiarly affected, so for that the reading would be unaltered. There is a slight additional effect nearer the equator due to the spinning earth - a spring balance there will register low, a see-saw balance won't be affected.
you would be closer to the equator!!
The apparent force of gravity on earth is not the same all over, the spin of the earth means that you weigh less at the equator than at the poles, due to the centripetal force from the earths spin. You will weigh about 0.3% less at the equator. If the earth spun faster still, this difference would be even more apparent.