That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
The vertical component of the initial velocity is 
The horizontal component of the initial velocity is 
The horizontal displacement when the object reaches maximum height is 
The given parameters;
the horizontal displacement of the object, = x
the vertical displacement of the object, = y
acceleration due to gravity, = g
time of motion, = t
The vertical component of the initial velocity is given as;

The horizontal component of the initial velocity is calculated as;

The time to reach to the maximum height is calculated as;

The horizontal displacement when the object reaches maximum height is calculated as;

Learn more here: brainly.com/question/20689870
Answer:
C. 441 N
Explanation:
Gravitational force between two objects can by calculated by the formula
= G m₁m₂ / r² , m₁ and m₂ are masses at distance r
= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²
= 44.09 x 10
= 440.9 N
= 441 N .
4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.
The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
So,
E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>