Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s
1 answer:
Answer:
Part A
Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;
= -
Where;
= The action force
= The reaction force
Part B
The law indicates that the force with which a rocket ship uses in taking off from the Earth, is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-)
Part C
The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space
Explanation:
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Answer:
A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.
Explanation:
A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.
If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:
vₓ = 16.6 m/s i
In the vertical direction, the football, once released, is in free fall, starting from rest.
So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:
vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s
Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:
vy = -17.2 m/s j
So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:
v = 16.6 m/s i -17.2 m/s j
b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:
v = 23.9 m/s
c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:
tg θ =
⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.
This is a uniform rectilinear motion (MRU) exercise.
To start solving this exercise, we obtain the following data:
<h3><u>Data:</u></h3>- v = 4.6 m/s
- d = ¿?
- t = 10 sec
To calculate distance, speed is multiplied by time.
We apply the following formula: d = v * t.
We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:
Therefore, the speed at 10 seconds is 46 meters.