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.
A desert on the equator would be an anomaly as there are no deserts on the equator.
In that case, your weight remains absolutely constant and does not budge one iota.
The weight of an object is slightly less at the equator than at the poles because of the earth's tilt on its axis.
delhi is near the equator. the earth is thicker at the equator, object is further away from earth than if it at pole. heavier at the pole.
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.
you would be closer to the equator!!
The zeroth latitude is at the equator.
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.
Because of centripetal acceleration you will weigh a tiny amount less at the equator than at the poles.
equator