Answer: a) speed = 3.45 × 10^-2 m/s
b) speed = 1.38 × 10^-1 m/s
Explanation: shown in the attachment
According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is -6048 N
<h3>
What does Newton's third law say ?</h3>
The law state that in every action, there will be equal and opposite reaction.
Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.
Let us first calculate the force involved in the acceleration of the rocket by using the formula
F = ma
Where mass m = 87.4 kg, acceleration a = 69.2 m/s2
Substitute the two parameters into the formula
F = 87.4 x 69.2
F = 6048.08 N
According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.
Therefore, the normal force acting on the astronaut is -6048 N approximately
Learn more about forces here: brainly.com/question/12970081
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B, is the answer to the question
Answer:
As the masses get larger the force gets larger
Explanation:
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

and since the electric field E in between parallel plates separated a distance d and under a potential difference
, is given by:

then :

We want to find when the particle reaches velocity zero via kinematics:

We replace this time (t) in the kinematic equation for the particle displacement:

Replacing the values with the information given, converting the distance d into meters (0.01 m), using
, and the electron's kinetic energy:

we get:
Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]