Q: On a scale drawing, \${1}/{4}\$ inch \$=1\$ foot. What are the dimensions on the scale drawing for a room that is \$18\$ feet by \$16\$ feet?

We are told in the scale drawing is:\$\dfrac{1}{4}\ inch=1\ feett\$To find how many inches to make the length of the room at \$18\$ feet multiply each side of teh equation by \$18\$\$18 \times \dfrac{1}{4}\ inch=18 \times 1\ feet\$\$\dfrac{18}{4}\ inches=18\ feet\$\$\dfrac{16+2}{4} inches=18\ feet\$\$\left( \dfrac{16}{4}+\dfrac{2}{4}\right) inches=18\ feet\$\$\left( 4+ \dfrac{1}{2}\right) \ inches =18\ feet\$\$4\dfrac{1}{2} \ inches =18\ feet\$To find how many inches to makes the width of the room at \$16\$ feet multiply each side of the equation by \$16\$\$16 \times \dfrac{1}{4}\ inches =16 \times 1\ feet\$\$\dfrac{16}{4} \ inches =16\ feet\$\$4\ inches =16\ feet\$To draw teh room to scale you would draw a rectangle:\$4\dfrac{1}{2}\ inches \$ by \$4\$ inches

Q: The distance \$r\$ of the block at time \$t\$ is :

Force on block along slot \$ = m \omega^{2} r = ma = m \left (\dfrac{v dv}{dr} \right ) \$ \$\displaystyle \int^{v}_{0} v dv = \int^{r}_{R/2} \omega^{2} rdr \$ On integrating we get \$ \dfrac{v^2}{2} = \dfrac{\omega^2}{2} \left (r^{2} - \dfrac{R^2}{4} \right ) \Rightarrow v = \omega \sqrt{r^{2} - \dfrac{R^2}{4}} = \dfrac{dr}{dt} \$ \$ \Rightarrow \displaystyle \int^{r}_{R/4} \dfrac{dr}{\sqrt{r^{2} - \dfrac{R^2}{4}}} = \int^{t}_{0} \omega dt \$ \$ \ln \left (\dfrac{r + \sqrt{r^{2} - \dfrac{R^2}{4}}}{\dfrac{R}{2}} \right ) - \ln \left (\dfrac{R/2 + \sqrt{\dfrac{R^2}{4} - \dfrac{R^2}{4}}}{\dfrac{R}{2}} \right ) = \omega t \$ \$ \Rightarrow r + \sqrt{r^{2} - \dfrac{R^2}{4}} = \dfrac{R}{2} e^{\omega t} \$ \$ \Rightarrow r^{2} - \dfrac{R^2}{4} = \dfrac{R^2}{4} e^{2 \omega t} + r^{2} - 2r \dfrac{R}{2} e^{\omega t} \$

Q: All the surfaces shown in figure (12-E15) are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through a length \$x_0\$ when the system is released. Find the time period(s) of the two simple harmonic motions.

At any position, let the velocities be \$v_{1}\$ and \$v_{2}\$ respectively.Here, \$v_{1}=\$ velocity of \$m\$ with respect to \$M\$.By energy methodTotal Energy = Constant\$(1/2)Mv^{2}+(1/2)m(v_{1}-v_{2})^{2}+(1/2)k(x_{1}+x_{2})^{2}=\$ Constant ..(i)[\$v_{1}-v_{2}=\$ Absolute velocity of mass \$m\$ as seen from the road.]Again, from law of conservation of momentum\$mx_{2}=mx_{1}\Rightarrow x_{1}=\dfrac{M}{m}x_{2}\$ \$....(1)\$\$mv_{2}=m(v_{1}-v_{2})\Rightarrow (v_{1}-v_{2})=\dfrac{M}{m}v_{2}\$ \$.....(2)\$Putting the above values in equation \$(1)\$, we get\$\dfrac{1}{2}Mv_{2}^{2}+\dfrac{1}{2}m\dfrac{M^{2}}{m^{2}}v_{2}^{2}+\dfrac{1}{2}kx_{2}^{2}\left(1+\dfrac{M}{m}\right)^{2}=\$ constant\$\therefore M\left(1+\dfrac{M}{m}\right)v_{2}+k\left(1+\dfrac{M}{m}\right)^{2}x_{2}^{2}=\$ Constant.\$mv_{2}^{2}+l\left(1+\dfrac{M}{m}\right)x_{2}^{2}=\$ constantTaking derivative of both sides,\$M\times 2v_{2}\dfrac{dv_{2}}{dt}+k\dfrac{(M+m)}{m}--ex_{2}^{2}\dfrac{dx_{2}}{dt}=0\$\$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0\$ [because, \$c_{2}=\dfrac{dx_{2}}{dt}\$]Taking derivative of both sides,\$M\times 2v_{2}\dfrac{dv_{2}}{}+k\dfrac{(M+m)}{m}-ex_{2}^{2}\dfrac{dx_{2}}{dt}=0\$\$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0\$ [because, \$v_{2}=\dfrac{dx_{2}}{dt}\$]\$\Rightarrow \dfrac{a_{2}}{x_{2}}=-\dfrac{k(M+m)}{Mm}=\omega^{2}\$\$\therefore \omega=\sqrt{\dfrac{k(M+m)}{Mm}}\$So, Time period, \$T=2\pi \sqrt{\dfrac{Mm}{k(M+m)}}\$

Question 1

SI unit of potential energy?

Accepted Answer

Please understand that the SI unit for potential energy  is the same as the SI unit for all energy, the Joule.The Joule in SI base units is defined as:$J = \dfrac{kg \cdot m^2}{s^2}$Please consider the equation for kinetic energy.$K.E. = \dfrac{1}{2}mV^2$The units for V is $\dfrac{m}{s}$Square that $\dfrac{m^2}{s^2}$Multiply by the units for mass:$\dfrac{kg \cdot m^2}{s^2}$Let's consider Gravitational potetial energy:$P.E. = mgh$The units for m is kg:$P.E. = kggh$The units for g is $\dfrac{m}{s^2}:$$P.E. = kg\dfrac{m}{s^2}h$The units for h is m:$P.E. = \dfrac{kg \cdot m^2}{s^2}$Lets look at $E = mC^{2}$The units for m is kg:$E = kgC^{2}$The units for C is $\dfrac{m}{s}$ square it $\dfrac{m^2}{s^2}$$E = \dfrac{kg \cdot m^2}{s^2}$I hope that this helps you to understand that the SI unit for all types of energy is, the Joule.

Question 2

How many feet are in a football field that is 100 yards?

Accepted Answer

Answer300 ftExplanation :Since 1 yard = 3 feet, we can multiply 100 yd by  $\dfrac{3 ft}{1 yd} $ We are essentially multiplying by 1, but doing so allows us to cancel out the yd unit so that we are left with only ft.$ \dfrac{100 yd}{1}.\dfrac{3 ft}{1 yd} =  300 ft $

Question 3

What is the unit of viscosity? Please explain.

Accepted Answer

The viscosity of a fluid is a measure of its resistance to flow.We usually think of the viscosity of a liquid as its &#34;thickness&#34;.For example, we say that honey is &#34;thicker&#34; (has a higher viscosity) than water.One way to measure viscosity is to place the fluid between two flat parallel plates and slide one plate past the other.In terms of the above diagram, the formula for viscosity &#951; is$  \eta = \frac { F \delta }{Av } $Thus the units of visocity are $ \frac { N \times m}{ m^2 \times  m . s^{-1} }= N.sm.m^{-2} = P a.s $If a fluid is placed between two  $ 1m^2 $ plates that are 1 m apart and one plate is pushed sideways with a force of 1 N and it moves at a velocity of $ 1 m/s^{-1} $ then the fluid has a viscosity $ \eta  = 1 Pa.s $Many chemists still use an old unit of viscosity, the poise (P), named after the French physician Jean Poiseuille.It is convenient unit because $ 1m Pa.s = 1cP $For example , the viscosity of water at $ 20^0 C $ is $  1.002 mPa.s $ or $ 1.002 cP $Honey has a viscosity of about $ 10 Pa.s $ or $ 10000cP $

Question 4

On a scale drawing, ${1}/{4}$ inch $=1$ foot. What are the dimensions on the scale drawing for a room that is $18$ feet by $16$ feet?

Accepted Answer

We are told in the scale drawing is:$\dfrac{1}{4}\ inch=1\ feett$To find how many inches to make the length of the room at $18$ feet multiply each side of teh equation by $18$$18 \times \dfrac{1}{4}\ inch=18 \times 1\ feet$$\dfrac{18}{4}\ inches=18\ feet$$\dfrac{16+2}{4} inches=18\ feet$$\left( \dfrac{16}{4}+\dfrac{2}{4}\right) inches=18\ feet$$\left( 4+ \dfrac{1}{2}\right) \ inches =18\ feet$$4\dfrac{1}{2} \ inches =18\ feet$To find how many inches to makes the width of the room at $16$ feet multiply each side of the equation by $16$$16 \times \dfrac{1}{4}\ inches =16 \times 1\ feet$$\dfrac{16}{4} \ inches =16\ feet$$4\ inches =16\ feet$To draw teh room to scale you would draw a rectangle:$4\dfrac{1}{2}\ inches $ by $4$ inches

Question 5

One year on Mercury is equal to $87.97$ Earth days. One year on Pluto is three times the length of one Mercury year minus $16.21$ days. How long is one year on Pluto?

Accepted Answer

This is a tricky one as there may be some ambiguity about $16.21$ days which is: to which planet is the day attributed? Also the units are tricky. They behave the same way numbers do!!!Assumption $1$From the sentence part &#34;of one Mercury year minus $16.21\ days&#34;$ I am assuming that the days are Mercury days. From $year$ minus $16.21\ days&#34;$ I am assuming that they are directly linked. Thus the days are directly attributable to the Mercury year.Assumption $2$Our annual cycle is split up into $365\ units$ of cycles in one solar year ( a day). The other planets will circle the sun at different speeds but will experience the equivalent effects our planet does. However, these will be at different rates. Thus for each planet their year may also be split up into $365$ rotational units.Building the initial equationBuilding the initial equationWe are talking about units of measurement in both days and years. So let the generic unit for day be $d$ and the generic unit for year be $y$. This gives us:Let the year unit of measurement for Mercury be $y_m$Let the day unit of measurement for Mercury be $d_m$Similarly for Earth we have $y_e$ and deye and $d_e$And for Pluto we have $y_p$ and $d_p$Another potential ambiguity.Consider the question in parts:&#34;One year on Mercury is $87.97$ on earth$\rightarrow 1 y_m=87.97 d_c...(1)$No doubt about teh meaning!&#34;One year on Pluto is $3$ times one year on Mercury minus $16.21$ days&#34;Does this mean: $1y_p=3(1 y_m-1621 d_m)$Or does it mean:$1y_p=3(1 y_m)-1621 d_m$Assuming this is the case: $1y_p=3(1 y_m-1621 d_m) ...(2)$Substitute $(1)$ in $(2)$ giving:$1y_b=3(87.97d_c- 16.21 d_m)...(3)$To convert $16.21 d_m$ into $d_c$From $(1)\ 1y_m=87.97d_c \rightarrow 365d_m =87.97d_c$Divide both sides by $365$ giving:$1d_m =\dfrac{87.97}{365} d_c$$1y_p=3 \left( 87.97d_c -\dfrac{16.21 \times 87.97}{365} d_c \right) ...(3_a)$I get $1y_o \approx 252.19d_c$ But we need this in yearsSo $1y_p \approx \dfrac{252.19}{365} y_c \approx 0.69 y_c$

Question 6

The dimensions of Planck's constant is

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Question 7

Choose the correct answer from the alternatives given.Time period is the time taken to complete one oscillation. This definition of time period is true for

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Question 8

The distance $r$ of the block at time $t$ is :

Accepted Answer

Force on block along slot $ = m \omega^{2} r = ma = m \left  (\dfrac{v dv}{dr}  \right  ) $ $\displaystyle \int^{v}_{0} v dv = \int^{r}_{R/2} \omega^{2} rdr $ On integrating we get $ \dfrac{v^2}{2} = \dfrac{\omega^2}{2} \left  (r^{2} - \dfrac{R^2}{4} \right  ) \Rightarrow v = \omega \sqrt{r^{2} - \dfrac{R^2}{4}} = \dfrac{dr}{dt} $ $ \Rightarrow \displaystyle \int^{r}_{R/4} \dfrac{dr}{\sqrt{r^{2} - \dfrac{R^2}{4}}} = \int^{t}_{0} \omega dt $ $ \ln \left  (\dfrac{r + \sqrt{r^{2} - \dfrac{R^2}{4}}}{\dfrac{R}{2}}  \right  ) - \ln \left  (\dfrac{R/2 + \sqrt{\dfrac{R^2}{4} - \dfrac{R^2}{4}}}{\dfrac{R}{2}} \right  ) = \omega t $ $ \Rightarrow r + \sqrt{r^{2} - \dfrac{R^2}{4}} = \dfrac{R}{2} e^{\omega t} $ $ \Rightarrow r^{2} - \dfrac{R^2}{4} = \dfrac{R^2}{4} e^{2 \omega t} + r^{2} - 2r \dfrac{R}{2} e^{\omega t} $

Question 9

How was it determined that a parsec is $ 3.26 $ light years?

Accepted Answer

Parallax is an effective way to measure distance to nearby stars because it relies on geometry. When astronomers use parallax, they are measuring how a star appears to move against its background. The unit parsec refers to the distance that an object would have to be from the Earth  to have a parallax angle of $ 1 $ arc-second.In the figure above, the angle, $ \alpha $ is the angle that is measured by the Earth on opposite sides of the sun. The parallax angle, $ p $ is half of this angle.$ p=\frac{1}{2}\alpha$If we defines $ p $ to be $ 1 $ arc-second, then our object will be $ 1 $ parsec away. Since light from the sun takes $ 8 $ minutes and $ 20 $ seconds to reach the Earth,. we know that;$ 1 AU=8.33 $ light minutesWe can use this information to convert our parsec into light years with the tangent formula.$ tan(p)=\frac{8.33 \quad light \quad minutes}{d}$Or;$ d=\frac{8.33 \quad light\quad minutes}{tan(p)}$If we convert $ p $  to radians, than we can use the small angle approximation, $ tan(\theta)\approx \theta$.$ 1 arc-second=4.85 \times 10^{-6} $ radiansPlugging this in for $ p $  and using the small angle approximation;$ d=\frac{8.33 \quad light\quad minutes}{4.85\times 10^{-6}}$=$ 1.72 \times 10^6 $ light-minutes=$ 3.26 $ light years

Question 10

All the surfaces shown in figure (12-E15) are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through a length $x_{0}$ when the system is released.
Find the time period(s) of the two simple harmonic motions.

Accepted Answer

At any position, let the velocities be $v_{1}$ and $v_{2}$ respectively.Here, $v_{1}=$ velocity of $m$ with respect to $M$.By energy methodTotal Energy = Constant$(1/2)Mv^{2}+(1/2)m(v_{1}-v_{2})^{2}+(1/2)k(x_{1}+x_{2})^{2}=$ Constant ..(i)[$v_{1}-v_{2}=$ Absolute velocity of mass $m$ as seen from the road.]Again, from law of conservation of momentum$mx_{2}=mx_{1}\Rightarrow x_{1}=\dfrac{M}{m}x_{2}$      $....(1)$$mv_{2}=m(v_{1}-v_{2})\Rightarrow (v_{1}-v_{2})=\dfrac{M}{m}v_{2}$     $.....(2)$Putting the above values in equation $(1)$, we get$\dfrac{1}{2}Mv_{2}^{2}+\dfrac{1}{2}m\dfrac{M^{2}}{m^{2}}v_{2}^{2}+\dfrac{1}{2}kx_{2}^{2}\left(1+\dfrac{M}{m}\right)^{2}=$ constant$\therefore M\left(1+\dfrac{M}{m}\right)v_{2}+k\left(1+\dfrac{M}{m}\right)^{2}x_{2}^{2}=$ Constant.$mv_{2}^{2}+l\left(1+\dfrac{M}{m}\right)x_{2}^{2}=$ constantTaking derivative of both sides,$M\times 2v_{2}\dfrac{dv_{2}}{dt}+k\dfrac{(M+m)}{m}--ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$ [because, $c_{2}=\dfrac{dx_{2}}{dt}$]Taking derivative of both sides,$M\times 2v_{2}\dfrac{dv_{2}}{}+k\dfrac{(M+m)}{m}-ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$  [because, $v_{2}=\dfrac{dx_{2}}{dt}$]$\Rightarrow \dfrac{a_{2}}{x_{2}}=-\dfrac{k(M+m)}{Mm}=\omega^{2}$$\therefore \omega=\sqrt{\dfrac{k(M+m)}{Mm}}$So, Time period, $T=2\pi \sqrt{\dfrac{Mm}{k(M+m)}}$

Basic Science

General Knowledge
5651 Views

SI unit of potential energy?

How many feet are in a football field that is 100 yards?

What is the unit of viscosity? Please explain.

On a scale drawing, $1 / 4$ inch $= 1$ foot. What are the dimensions on the scale drawing for a room that is $18$ feet by $16$ feet?

One year on Mercury is equal to $87.97$ Earth days. One year on Pluto is three times the length of one Mercury year minus $16.21$ days. How long is one year on Pluto?

The dimensions of Planck's constant is

Choose the correct answer from the alternatives given.
Time period is the time taken to complete one oscillation. This definition of time period is true for

The distance $r$ of the block at time $t$ is :

How was it determined that a parsec is $3.26$ light years?

Basic Science

General Knowledge 5651 Views

SI unit of potential energy?

How many feet are in a football field that is 100 yards?

What is the unit of viscosity? Please explain.

On a scale drawing, 1/4 inch =1 foot. What are the dimensions on the scale drawing for a room that is 18 feet by 16 feet?

One year on Mercury is equal to 87.97 Earth days. One year on Pluto is three times the length of one Mercury year minus 16.21 days. How long is one year on Pluto?

The dimensions of Planck's constant is

Choose the correct answer from the alternatives given. Time period is the time taken to complete one oscillation. This definition of time period is true for

The distance r of the block at time t is :

How was it determined that a parsec is 3.26 light years?

General Knowledge
5651 Views

On a scale drawing, $1 / 4$ inch $= 1$ foot. What are the dimensions on the scale drawing for a room that is $18$ feet by $16$ feet?

One year on Mercury is equal to $87.97$ Earth days. One year on Pluto is three times the length of one Mercury year minus $16.21$ days. How long is one year on Pluto?

Choose the correct answer from the alternatives given.
Time period is the time taken to complete one oscillation. This definition of time period is true for

The distance $r$ of the block at time $t$ is :

How was it determined that a parsec is $3.26$ light years?