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2 years ago

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A rocket is launched from Earth (mass ME, radius RE) with velocity v° and reaches radial distance r=6RE with velocity v°/10. Exp

ress V° in terms of ME RE. What is the maximum height that the rocket could reach if launched vertically.

1 answer:

Aliun [14]2 years ago

5 0

The velocity, expressed in terms of ME, RE is given as $V_0 = \sqrt{98.99R_E}$ .

The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.

<h3>Maximum height of the rocket</h3>

The maximum height reached by the rocket can be modeled using conservation of energy as shown below;

P.Ei + K.Ei = K.Ef + P.Ef

$M_EgR_E + \frac{1}{2} M_EV_0^2= \frac{1}{2} M_E(\frac{V_o}{10} )^2+ M_Eg(6R_E)\\\\\frac{1}{2} M_EV_0^2 - \frac{1}{2} M_E(\frac{V_o}{10} )^2 = M_Eg(6R_E) - M_EgR_E\\\\0.495M_EV_0^2= 5gM_ER_E\\\\0.495V_0^2= 5gR_E\\\\V_0 = \sqrt{\frac{5gR_E}{0.495} } \\\\V_0 = \sqrt{98.99R_E}$

<h3>Maximum height when it is launched vertically</h3>

P.E = K.E

mgH = ¹/₂mv²

H = (¹/₂v₀²)/g

Learn more about conservation of energy here: brainly.com/question/166559

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

STatiana [176]

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

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3 years ago

iogann1982 [59]

The correct answer is:
<span>C: in the protons and neutrons of an atom

In fact, the nuclear energy refers to the binding energy of the nucleons (protons and neutrons) of an atom. The protons and the neutrons are held together by the strong nuclear interaction, one of the four fundamental forces of nature, and the energy associated to this interaction is called nuclear energy.

</span>

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2 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

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melamori03 [73]

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bija089 [108]

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