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pishuonlain [190]
3 years ago
11

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

garding any air resistance, what maximum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s 2 .
Physics
1 answer:
vredina [299]3 years ago
3 0

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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Answer:

d = 44.64 m

Explanation:

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a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

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v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

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Explanation:

La fuerza ejercida entre las dos partículas se calcula por la Ley de Coulomb (The force exerted between the two particles is determined by Coulomb's Law):

F = k \cdot \frac{q_{1}\cdot q_{2}}{r^{2}}

Donde (Where):

k - Constante electrostática, medido en \frac{N\cdot m^{2}}{C^{2}} (Electrostatic constant, measured in \frac{N\cdot m^{2}}{C^{2}}).

q_{1}, q_{2} - Magnitudes de las cargas de cada partícula, medidos en Coulombs. (Magnitudes of charges from each particle, measured in Coulombs).

r - Distancia entre las partículas, medida en metros.

La fuerza electrostática es (Electrostatic force is):

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \frac{(5\times 10^{-6}\,C)\cdot (5\times 10^{-6}\,C)}{(0.20\,m)^{2}}

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