Gravitational pull must be measured in units of _______.

Gravitational force must be measured in units of : meters per second squared In every Physics equation, usually the amount of Gravitational force is standardized and will be exactly the same as long as the event happens on the earth, which is 9.8 m/s^2.

Answered by Greenleafable | 2024-06-11

The direction of the** pull **is -70.93°, or approximately 71° South of West.
Given the vector as F= (−2980.0i^ + 8200.0j^)N
The magnitude of the pull is given by the formula below:
Magnitude of pull, |F| = √(F_x^2 + F_y^2)
Here, F_x is the force in the x-direction and F_y is the **force **in the y-direction. We know that:
F_x = -2980.0 N and F_y = 8200.0 N
Therefore, the magnitude of the pull is:
|F| = √(F_x^2 + F_y^2)|F| = √((-2980.0)^2 + (8200.0)^2)|F| = √(88840000) |F| = 9425.89 N
The direction of the pull can be found as follows:
θ = tan⁻¹(F_y/F_x)θ = tan⁻¹(8200.0/-2980.0)θ = -70.93°
The direction of the pull is -70.93°, or approximately 71° South of West.
learn more about **pull **on:
https://brainly.com/question/174980

Answered by sachinkaushik7700 | 2024-06-17

Gravitational pull is measured in units of Newtons (N), which represent the force of attraction between two masses. This measurement is derived from the basic SI units of mass, length, and time. The formula F = G ⋅ m 1 ​ ⋅ m 2 ​ / r 2 is used to compute this force, with G being the gravitational constant.
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Answered by Greenleafable | 2024-12-26