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Nataly [62]
3 years ago
8

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

The engines then fire, and the rocket accelerates up- ward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 29.80 m/s2. (a) For what time inter- val is the rocket in motion above the ground?
Physics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

y_{i}=0m

v_i=80\frac{m}{s}

a=4\frac{m}{s^2}

y_{e}=1000m

g=9.8\frac{m}{s^2}

From Kinematics, the position y as a function of time when the engine still works will be:

y(t)=v_it+\frac{1}{2}at^2

At what time the altitud will be y_{e}=1000m?

v_it+\frac{1}{2}at^2=y_{e} ⇒ \frac{1}{2}at^2+v_it-y_{e}=0

Using the quadratic formula: t_1=10s.

How much time does it take for the rocket to touch the ground? No the function of position is:

y(t)=y_{e}+v_et-\frac{1}{2}gt^2

Where our new initial position is y_{max}, the velocity when the engine breaks is v_e=v_i+at=120\frac{m}{s} and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

y_{e}+v_et-\frac{1}{2}gt^2=0

Again, using the quadratic ecuation:

t_2=24.49s

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

t_T=t_1+t_2=34.49s.

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0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for
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Which of the following is not a source of light? *<br> a) sun<br> b) star<br> c) mirror<br> d)cfl
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A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i
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Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k
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Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

B= magnetic field = 6.90×10-5 T.

We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

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Then from the expression above

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E = 6.90 x 10^-5 x 12388.32 V

E =0.855V

Hence, the average induced voltage between the tips of the wings is =0.855V

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3 years ago
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