Answer:
The distance covered by the rocket after fuel ran out is 
Explanation:
Given that the rocket moves with an acceleration 
time 
Since the rocket starts from rest initial velocity 
The distance it travelled within this time is given by

Velocity at this point is given by 

Given that at this height it runs out of fuel but travels further. Here final velocity
(maximum height), initial velocity
and time to zero velocity 
Thus it travels
more after fuel running out. The distance covered during this period is given

The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:

Learn more about electric field:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
Answer:
<h2>17.1 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
3800 g = 3.8 kg
We have
force = 3.8 × 4.5
We have the final answer as
<h3>17.1 N</h3>
Hope this helps you
Answer:
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