Solve this question plz

A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

geniusboy [140]

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

5 0

3 years ago

A kitten has a mass of 0.8 kg. It is moving forward with a momentum of 0.5 kg • m/s. What is the kitten’s velocity?

morpeh [17]

The kitten's velocity is V=0.625 m/s

<u>Explanation:</u>

<u>Solving the problem,</u>

Given

Mass=0.8 kg

Momentum=0.5 kg.m/s

Velocity=?

We have the formula,

P=M*V

V=P/M

V=0.5 kg.m/s/0.8 kg

V=0.625 m/s

The kitten's velocity is V=0.625 m/s

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5 0

4 years ago

While a kettle boils, 0.018kg of water changes to steam

Bumek [7]

As the water boils at a certain temperature, phase change happen without change in its temperature. The heat associated is called the latent heat of vaporization. We obtain the heat required by multiplying the mass of the water to the latent heat of vaporization.

Heat = 0.018 x <span>2.3 x 10000000 = 41400 J</span>

5 0

4 years ago

Which of the following is an example of qualitative data

NemiM [27]

Qualitative data is easy to explain. You either agree or disagree in writing. Wrong. You agree, disagree, or qualify. Meaning agree to somewhat extend, but not entirely. So you can agree that people should walk instead of talk in the progressive manner, but not in the impulsive manner.

3 0

4 years ago

A solenoid that is 66.1 cm long has a cross-sectional area of 13.8 cm2. There are 1410 turns of wire carrying a current of 8.01

Ymorist [56]

Answer:

a) Energy density of the magnetic field, u = 183.46 J/m³

b) Total energy, E = 0.167 J

Explanation:

a) Number of turns in the solenoid, N = 1410

Area, A = 13.8 cm² = 0.00138 m²

Current, I = 8.01 A

Length of the solenoid, l = 66.1 cm = 0.661 m

Energy density, u is given by the formula $u = \frac{B^2}{2 \mu_{0} }$

Where B is the magnetic field

The magnetic field in a solenoid is given by the formula, $B = \frac{N \mu_{0} I }{l}$

$B = \frac{1410 * 8.01* \mu_{0} }{0.661}$

$B = 17086.38 \mu_0$ T

$u = \frac{(17086.38 \mu_0)^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0}{2 }\\u = \frac{291944527.29 * 4\pi * 10^{-7} }{2 }\\u = 183.46 J/m^3$

b) The total energy = Energy density * Volume

E = u V

Volume = Area * Length

V = Al = 0.00138 * 0.661

V = 0.00091218 m³

E = 183.46 * 0.00091218

E = 0.167 J

4 0

3 years ago