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4 years ago

The more particles a substance has at a given temperature, the more thermal energy it has.

1 answer:

5 0

The statement '<span>The more particles a substance has at a given temperature, the more thermal energy it has' is true. </span><span>The kinetic molecular theory of gases has three main laws and one of them is the average kinetic energy of the particles in a gas. The average kinetic energy of the gas particles is the behavior and movement it does in the surroundings. It is directly proportional to temperature wherein if you increase the temperature, the kinetic energy of a particle also increases. It will also decrease its movement or its kinetic energy if the temperature lowers. </span>

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One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R

Tanzania [10]

(a) $24.6 m/s^2$

At a distance r=R from the centre of the planet, there is no effect due to the outer shell: so, the gravitational field strength at r=R is only determined by the gravity produced by the core of the planet.

So, the strength of the gravitational field is given by

$g= \frac{GM}{R^2}$

where

G is the gravitational constant

M = 6.24 × 10^24 kg is the mass of the core of the planet

R = 4.11 × 10^6 m is the radius of the core

Substituting into the equation, we find

$g= \frac{(6.67\cdot 10^{-11})(6.24\cdot 10^{24} kg)}{(4.11\cdot 10^6 m)^2}=24.6 m/s^2$

(b) $13.7 m/s^2$

at distance r=3R from the centre, the particle feels the effect of gravity due to both the core of the planet and the outer shell between R and 2R.

So, we have to consider the total mass that exerts the gravitational attraction at r=3R, which is the sum of the mass of the core (M) and the mass of the shell (4M):

M' = M + 4M = 5M

Therefore, the gravitational acceleration at r=3R will be

$g'= \frac{G(5M)}{(3R)^2}=\frac{5}{9}\frac{GM}{R^2} = \frac{5}{9}g$

And susbstituting

g = 24.6 m/s^2

found in the previous part, we find

$g' = \frac{5}{9} (24.6 m/s^2)=13.7 m/s^2$

6 0

3 years ago

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

4 years ago

Using this formula vf =vi+at if a vehicle starts from rest and. Accelerates forward at 4.5m/s for 8s what is the final velocity

iragen [17]

Answer: vf= 36 m/s

Explanation:

8 0

4 years ago

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago

Recall the raisin cake model of the universe. Our universe is expanding between the galaxies. You measure the recession velocity

Rina8888 [55]

Explanation:

Recession velocity of a galaxy is related to the distance at which the galaxy is located. This relationship is given by the hubble constant, as follows:

$v_r=HD$

Hubble constant is aproximate $70\frac{km/s}{Mpc}$ and 1 megaparsec (Mpc) is 3.26*10^6 light years. Rewriting for D:

$D=\frac{v_r}{H}$

For galaxy A:

$D=\frac{2000\frac{km}{s}}{70\frac{km/s}{Mpc}}\\\\D=28.57Mpc\\\\D=28.57Mpc*\frac{3.26*10^6ly}{1Mpc}=9.31*10^7ly$

For galaxy B:

$D=\frac{6000\frac{km}{s}}{70\frac{km/s}{Mpc}}\\\\D=85.71Mpc\\\\D=85.71Mpc*\frac{3.26*10^6ly}{1Mpc}=2.79*10^8ly$

5 0

3 years ago