What are the primary colors?

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

(c) In the first part we have:

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

3 years ago

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

4 0

2 years ago

The change in momentum (and impulse) are still 3000 N*s. Safety devices (like airbags, seatbelts and crumple zones) cause crashe

aleksklad [387]

Answer:

F = 2000 N

Explanation:

According to Newton's Second Law of Motion, the rate of change of momentum is equal to the force applied. Therefore, the formula written below will be used to solve this problem:

$F = \frac{\Delta P}{t}\\\\$