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Zina [86]
3 years ago
6

A rocket is fired from rest from the ground (y = 0) at time t0 = 0 s. As the rocket is burning its fuel, it moves vertically upw

ard with a constant acceleration of 3.25 m/s2 . At t1 = 10.0 s, all the fuel has been used up and the rocket is in free fall. Air resistance can be neglected up until t3 = 17.5 s when the rocket deploys a parachute. At this time the rocket immediately reaches its terminal velocity; this means that the rocket is no longer accelerating.
1a. (10 points) Find the maximum height of the rocket with respect to the ground. How long after being launched does it take for the rocket to reach this height?
1b. (10 points) How long after being launched does it take for the rocket to return to the ground? What is the rocket’s displacement over this entire trip? What is the rocket’s average velocity over this entire trip? What is the rocket’s average speed over this entire trip? What is the rocket’s average acceleration over this entire trip? Be sure to include both magnitude and direction for any vector quantities.
1c. (10 points) What is the rocket’s displacement from ta = 12.0 s to tb = 20.0 secs? What is the rocket’s average velocity from ta to tb? What is the rocket’s average speed from ta to tb? What is the rocket’s average acceleration from ta to tb? Be sure to include both magnitude and direction for any vector quantities.
1d. (10 points) Sketch a y v. t plot for the motion of the rocket. Be sure to label the positions of the rocket at t0 = 0 s, t1 = 10.0 s, t2 (when the rocket is at its highest point), t3 = 17.5 s, and t4 (when the rocket returns to the ground).

Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

Explanation:

attached is the solution

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sineoko [7]
W = mg = 350 newton
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3 years ago
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Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

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Answer:

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Explanation:

ac = v^2/r

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zysi [14]
<span>The support force is 15 N. The net force on the book is zero.</span>
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