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3 years ago

Attempt 3 of 2

Physics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

mass.

Explanation:

other physical factors are changeable but the mass of a particular substance is always constant.

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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:

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

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

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

You jump off a platform 134 m above the Neuse River. After you have free-fallen for the first 40 m, the bungee cord attached to

omeli [17]

Answer:

The acceleration at lowest point is 19.62 m/s^2

Explanation:

Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.

Lets Assume

Constant of string is K

By using the conservation of energy we will have the following equation

1/2 x 80^2 x K = m x 9.81 x 120

3200 K = 1177.2 m

K = 1177.2 m / 3200

K = 0.368 m

At the lowest point we will have

a = ( K x X - m x g ) / m

a = ( 0.368 m x 80 - m x 9.81 ) / m

a = 19.62 m / s^2

So, the acceleration at lowest point is 19.62 m/s^2

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A 3 kg model airplane is traveling at a speed of 33 m/s. The operator then increases the speed up to 45 m/s in 2 seconds. How mu

nlexa [21]

Answer:

Force = 18N

Explanation:

Force = [ mass ( final velocity - initial velocity ) ] / time taken

using the formula, here mass is 3 kg, final velocity = 45 m/s , initial velocity = 45 m/s , time taken = 2 seconds

Force = [ 3 ( 45 - 33 ) ] / 2

Force = 18N

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1,3,4 I’m pretty sur e

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A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours

fredd [130]

30km. 24 the first two hours and 6 the half hour

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