You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.
Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.
Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.
Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.
Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.
Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
The weight of an object is slightly less at the equator than at the poles because of the earth's tilt on its axis.
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.
More information is needed.
First you have to convert weight into mass. This is dependent on the acceleration the mass is experiencing (either gravitational or centrifugal). If it is gravitational and it is at or near the surface of the Earth then mass=weight/9.81m/s2 If it is centrifugal then a=v2/r and mass=weight*r/v2 Then to find momentum just multiply mass by velocity.
Mass. . . . . same at the poles as it is at the equator. Weight . . . more at the poles Cost . . . . . more at the poles
a mass/weight unit cannot be converted into rpm, because rpm is unit of angular velocity.
Depends entirely on the caliber, weight and velocity of the projectile along with the weight of the firearm
GRAVITYThe acceleration due to gravity is a force related to Earth's mass and is not dependent on its rotation - gravity would not change if the Earth ceased to rotate. WEIGHT However, if the Earth ceased to rotate, someone standing on the equator would weigh more - this increase in weight effect would decrease as you moved the person to the poles to do the comparison.
There are two reasons, both related to the Earth's rotation. At the equator, the Earth and the objects on it are at their maximum rotational velocity (about 465 meters per second). This causes the surface of the Earth at the equator to bulge farther from the center of mass, by an average of about 3.5 kilometers.So the effect of gravity is higher at the North Pole than the equator because of:1) the rotational velocity acting to reduce the downward acceleration2) the greater average distance from the Earth's center at the equator, since gravity decreases with distance from the center of mass.The difference, however, is only 0.05%, or 1/200 of the weight.
No, velocity is by definition just a rate of movement that is a scalar quantity.
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.
Other things (the volume and shape) being equal, a greater weight would cause a greater terminal velocity.
Depdends on weight of gun, caliber, load and velocity of projectile
only earth. __________________ No. Adding weight will slow it down, but the angular momentum would be preserved. In order to stop it completely, some opposite force would need to be exerted to neutralize the angular momentum.
That is called terminal velocity.That is called terminal velocity.That is called terminal velocity.That is called terminal velocity.
yes...