According to this equation
F = G × m₁*m₂ ÷ r²
other than the mass, the distance also affects the gravitational force between two objects (same mass or not).
Therefore the correct answer is B. The distance between the objects
Future note* use formulas to help you figure these sort of questions out. (if they have a formula to begin with).
Answer:
(a) 6650246.305 N/C
(b) 24150268.34 N/C
(c) 6408227.848 N/C
(d) 665024.6305 N/C
Explanation:
Given:
Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]
Total charge of the ring (Q) = 75.0 μC = 
[1 μC = 10⁻⁶ C]
Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:
Plug in the given values for each point and solve.
(a)
Given:
, 
Electric field is given as:
(b)
Given:
, 
Electric field is given as:
(c)
Given:
, 
Electric field is given as:
(d)
Given:
, 
Electric field is given as:
Based on the calculations, the angle through which the tire rotates is equal to 4.26 radians and 244.0 degrees.
<h3>How to calculate the angle?</h3>
In Physics, the distance covered by an object in circular motion can be calculated by using this formula:
S = rθ
<u>Where:</u>
- r is the radius of a circular path.
- θ is the angle measured in radians.
Substituting the given parameters into the formula, we have;
1.87 = 0.44 × θ
θ = 1.87/0.44
θ = 4.26 radians.
Next, we would convert this value in radians to degrees:
θ = 4.26 × 180/π
θ = 4.26 × 180/3.142
θ = 244.0 degrees.
Read more on radians here: brainly.com/question/19758686
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The popular GPS devices that people use to find directions while driving use "Global Navigation Satellite System (GNSS)".
<u>Explanation:</u>
The umbrella term for all global satellite tracking systems is GNSS i.e Global Satellite Navigation System. This involves satellite constellations circulating over the surface of the earth and continuous signal transmission that allow users to evaluate their location.
A satellite array of 18–30 medium Earth Orbit (MEO) satellites distributed across several orbital planes typically achieves greater coverage for each network. The specific systems differ, but use > 50 ° orbital inclinations and approximately twelve hours orbital cycles.
