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3 years ago

A space rocket is launched and accelerates uniformly to 160 m/s in 4.5 s. Calculate the acceleration of the rocket.

Physics

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

$a = \frac{v}{t} \\$

v is the velocity

t is the time

From the question we have

$a = \frac{160}{4.5} = 35.55555...$

We have the final answer as

Hope this helps you

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Nick added five yards to the maximum distance that he can throw a football which component has he most likely added to his train

aleksandr82 [10.1K]

I believe the answer is: C.)weight lifting twice a week

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3 years ago

A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation change

irina [24]

Answer:

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes will be equal.

Explanation:

In principle we know the following:

<u>Internal Energy:</u> is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as $U=mC_{V}\Delta T$ where $m$ is the mass (kg), $C_{V}$ is a specific constant-volume (kJ/kg°C) and $\Delta T$ is the Temperature change in °C.

<u>Kinetic Energy:</u> denotes the work done on an object (of given mass $m$ ) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as $K=\frac{1}{2}mv^2$ where $v$ is the velocity of the object in (m/s).
<u>Potential Energy:</u> denotes the energy occupied by an object (of given mass $m$ ) due to its position with respect to another object. In S.I. is defined as $P=mgh$ , where $g$ is the gravity constant equal to $9,81m/s^2$ and $h$ is the elevation (meters).

<em>Note: The Internal energy is unaffected by the Kinetic and Potential Energies.</em>

<u>Given Information:</u>

Temperature Change 0°C → 18°C ( thus $\Delta T=18$ °C )
Object velocity we shall call it $v_{o}$ and $v_{f}$ , for initial and final, respectively. Here we also know that $v_{o}=0m/s^2$
Object elevation we shall call it $h_{o}$ and $h_{f}$ , for initial and final, respectively. Here we also know that $h_{o}= 0m$

∴<em> We are trying to find $v_{f}$ and $h_{f}$ of the air where $U$ , $K$ and $P$ are equal.</em>

Lets look at the change in Energy for each.

<u>Step 1: Change in Kinetic Energy=Change in Internal Energy</u>

$\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T$

Here we recall that $v_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T$

$\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}$ Eqn(1)

<u>Step 2: Change in Potential Energy=Change in Internal Energy</u>

$\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T$

Here we recall that $h_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T\\$

$h_{f}=\frac{C_{V}\Delta T}{g}$ Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

$v_{f}=\sqrt{2*718*18}=160.77m/s$

$h_{f}=\frac{718*18}{9.81}=1,317.43m$

Thus we can conclude that for the air final velocity $v_{f}=160.77m/s$ and final elevation $h_{f}=1,317.43m$ the internal, kinetic, and potential energy changes will be equal.

3 0

3 years ago

Put the following items in order of the scientific method:Draw conclusions. Ask a question. Develop a theory. Analyze data. Make

muminat

Ask a question
Make an observation
Develop a hypothesis
Experiment
Analyse data
Develop a theory
Hope this helped. :)

6 0

3 years ago

True or false. The number of images formed depends upon the angle between the mirrors.

Gre4nikov [31]

Explanation:

The number of images formed by two adjacent plane mirrors depends on the angle between the mirror.

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4 years ago

A space rocket is launched and accelerates uniformly to 160 m/s in 4.5 s. Calculate the acceleration of the rocket.​

A space rocket is launched and accelerates uniformly to 160 m/s in 4.5 s. Calculate the acceleration of the rocket.