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## Using mirror formula how you will prove that ''when an object is placed between C and F, it forms its image between C and ∞"?

A08
Featured 6 months ago

Assuming that the question is about spherical mirrors and has been posted under flat mirror by oversight only.

We know that Mirror formula is the relationship between object distance $u$, image distance $v$ and focal length $f$ of the mirror and is given as below*.

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .....(1)

A. When Object is placed at $\text{C}$
Object distance $u = 2 f$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{2 f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{2 f}$
$\implies \frac{1}{v} = \frac{1}{2 f}$
$\implies v = 2 f$
Implies that image is formed at $\text{C}$

B. Object is placed at $\text{F}$
Object distance $u = f$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{f} = 0$
$\implies v = \infty$
Implies that image is formed at $\infty$

C. Object is placed between $\text{C and F}$
Object distance $u = n f$, where $n$ lies between $1 \mathmr{and} 2$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{n f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{n f}$
$\implies \frac{1}{v} = \frac{n - 1}{n f}$
$\implies v = \frac{n f}{n - 1}$
Inserting any value of $n$ between $1 \mathmr{and} 2$, we see that image is formed between $\text{C} \mathmr{and} \infty$.

Note that for $n = 1$ we have case number A and for $n = 2$ we get case number B as discussed above.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

*Sign Convention: For spherical mirrors, normally object distance is taken as negative, as it is measured against direction of light.Image distance is positive for real images and is in the direction of travel of light. Focal length is positive accordingly for a converging mirror as it is measured in the direction of travel of light.
If sign convention is followed then Mirror Formula for a converging mirror becomes

$\frac{1}{u} + \frac{1}{f} = \frac{1}{v}$

and calculations are to be done appropriately.

## Do mass influence weight?

Morgan
Featured 6 months ago

Yes.

#### Explanation:

Mass directly influences weight. The weight of a stationary object is equal to the force of gravity acting upon it, where ${\vec{F}}_{g} = m g$.

A basic force diagram for a stationary object with no outside forces:

Where $\vec{n}$ is the normal force.

The normal force is the force exerted by a surface on the object which presses down on the surface. For example, when you stand on a scale, it measures your weight by measuring the normal force-the force exerted upwards by the scale on you to support your weight, which is equal in magnitude to the force you exert downwards on the scale (due to gravity).

You can see that the normal force is equal and opposite to the force of gravity in this situation, yielding a net force statement of:

$\sum \vec{F} = {\vec{F}}_{n e t} = \vec{n} - {\vec{F}}_{g} = \vec{0}$

$\vec{n} = {\vec{F}}_{g} = m g$

For the weight of an object which is accelerating, the net force is no longer zero.

$\vec{n} - {\vec{F}}_{g} = m \vec{a}$

And therefore,

$\vec{n} = m \vec{a} + {\vec{F}}_{g}$

$\implies \vec{n} = m \vec{a} + m g$

$\implies \vec{n} = m \left(\vec{a} + g\right)$

In both cases, we see that the mass of an object is directly related to its weight.

## Explain kinetic energy and it derivation?

A08
Featured 4 months ago

Energy possessed by an object due to its motion is termed as its Kinetic Energy.

It follows that if there is no motion, or the object is stationary, its kinetic energy is zero.

From Law of conservation of Energy we also know that energy can not be created. It however, can be changed from one form to another.

Derivation.
It is assumed mass is constant.
We know that work done in accelerating an object of mass $m$ through an infinitesimal time interval $\mathrm{dt}$ is given by the dot product of force $\vec{F}$ and infinitesimal displacement $\mathrm{dv} e c r$. Therefore total work done is
$W = \int \vec{F} \cdot \mathrm{dv} e c r$
Using Newton's Second law of Motion
$\vec{F} = m \vec{a}$
the integral becomes

$W = \int m \vec{a} \cdot \mathrm{dv} e c r$
Writing acceleration in terms of velocity $\vec{v}$ we get
$W = m \int \frac{\mathrm{dv} e c v}{\mathrm{dt}} \cdot \mathrm{dv} e c r$
Rearranging we get
$W = m \int \frac{\mathrm{dv} e c r}{\mathrm{dt}} \cdot \mathrm{dv} e c v$
We know that $\frac{\mathrm{dv} e c r}{\mathrm{dt}} \equiv \vec{v}$, therefore our integral becomes
$W = m \int \vec{v} \cdot \mathrm{dv} e c v$

Assuming that the object was at rest at time $t = 0$, we integrate from $t = 0$ to $t = t$ as the work done by the force to bring the object from rest to velocity $v$ is equal to the work necessary to reverse the process.
$W = m \int \vec{v} \cdot \mathrm{dv} e c v$
$W = m \left[\frac{1}{2} {v}^{2} + C\right]$
where $C$ is constant of integration.

As work done is zero when $v = 0$, $\implies C = 0$
Hence, $W = K E = \frac{1}{2} m {v}^{2}$

## The gravitational field at a point on the axis of a uniform disc, (mass = #M#, radius = #a#) forming #θ# at the point, is? (A)# ((GM)/a)cosθ# (B) #((GM)/a)(1-cosθ)# (C) #(2(GM))/a^2(1-cosθ)# (D) #(2(GM))/a^2(1-sinθ)#

A08
Featured 2 months ago

(C)

#### Explanation:

Figure 1

Let $P$ be the point on the axis of a uniform disc of mass $M$ and radius $a$, making an angle $\theta$ as shown in the figure above.

Let us consider an infinitesimal ring of width $\mathrm{dr}$ located at a distance $r$ from the centre of disc and of mass
#dm=(2πrdr)(M/(πa^2))# $\text{ } \left(M a s s = a r e a \times \mathrm{de} n s i t y\right)$
$\mathrm{dm} = \frac{2 M}{a} ^ 2 r \mathrm{dr}$ ......(1)

As shown in the figure below, we see that for an elemental part at $A$ of the ring, Gravitational field at point $P = \mathrm{dv} e c {E}_{G}$.

All points on the ring are equidistant form point $P$. Hence, the magnitude of gravitational field due to any of the elemental mass of ring is same at $P$

If we resolve this gravitational field in directions along the axis and perpendicular to the plane $O A P$, we see that for each element on the ring there exists an element located diametrically opposite. Vertical component of gravitational field due to the pair is zero. Therefore, vertical components due to complete ring vanish due to symmetry.

Hence axial components for ring in figure 1 from Newton's law of Universal Gravitation are
#dvecE_G=−(Gdmcosθ_e)/((r^2+x^2))#, along the axis
where #θ_e# is angle subtended by the infinitesimal ring at $P$ and is given as
#cosθ_e=x/sqrt(r^2+x^2)#

Inserting value from (1) and integrating over $r = 0 \text{ to } r = a$ we get

#|vecE_G|=∫_0^aGx((2M)/a^2rdr)/(r^2+x^2)^(3/2)#
#=>|vecE_G|=∫_0^a(2GMx)/a^2(r)/(r^2+x^2)^(3/2)dr#
Using standard integral we get
$\implies | {\vec{E}}_{G} | = \frac{2 G M x}{a} ^ 2 {\left[- \frac{1}{{r}^{2} + {x}^{2}} ^ \left(\frac{1}{2}\right)\right]}_{0}^{a}$

$\implies | {\vec{E}}_{G} | = \frac{2 G M x}{a} ^ 2 \left(- \frac{1}{\sqrt{{a}^{2} + {x}^{2}}} + \frac{1}{x}\right)$
$\implies | {\vec{E}}_{G} | = \frac{2 G M}{a} ^ 2 \left(1 - \frac{x}{\sqrt{{a}^{2} + {x}^{2}}}\right)$
$\implies | {\vec{E}}_{G} | = \frac{2 G M}{a} ^ 2 \left(1 - \cos \theta\right)$

## A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Truong-Son N.
Featured 1 month ago

I got $\text{9.93 m}$ horizontally from the starting horizontal position, with a maximum height of $\text{8.60 m}$ and a diagonal displacement of $\text{13.14 m}$.

We assume or know:

• Horizontal acceleration is ${\vec{a}}_{x} = 0$, since the projectile is let go after its launch (free horizontal motion).
• Starting position is ${\vec{x}}_{i} = 0 , {\vec{y}}_{i} = 0$, on the ground.

Defining up as positive and right as positive, the coordinates along the parabolic arc are given by the system of equations, assuming no air resistance:

${\vec{y}}_{f} = \frac{1}{2} \vec{g} {t}^{2} + {\vec{v}}_{i y} t + {\cancel{{\vec{y}}_{i}}}^{0}$
${\vec{x}}_{f} = {\cancel{\frac{1}{2} {\vec{a}}_{x} {t}^{2}}}^{0} + {\vec{v}}_{i x} t + {\cancel{{\vec{x}}_{i}}}^{0}$

The "distance from the starting point" is $\Delta \vec{x} \equiv {\vec{x}}_{f}$. For the diagonal distance, then

$\vec{d} = \sqrt{{\left(\Delta \vec{x}\right)}^{2} + {\left(\Delta \vec{y}\right)}^{2}}$. $\text{ "" } \boldsymbol{\left(1\right)}$

After our assumptions, here is what we don't know in red:

$\textcolor{red}{{\vec{y}}_{f}} = \frac{1}{2} \vec{g} {t}^{2} + {\vec{v}}_{i y} t$ $\text{ "" } \boldsymbol{\left(2\right)}$
$\textcolor{red}{{\vec{x}}_{f}} = {\vec{v}}_{i x} t$ $\text{ "" "" "" "" } \boldsymbol{\left(3\right)}$

We can find ${\vec{v}}_{i x}$ and ${\vec{v}}_{i y}$ using trigonometry; since $\cos \theta = \frac{{\vec{v}}_{x}}{\vec{v}}$ and $\sin \theta = \frac{{\vec{v}}_{y}}{\vec{v}}$:

${\vec{v}}_{i x} = \text{15 m/s" cdot cos(pi/3) = "7.5 m/s}$
${\vec{v}}_{i y} = \text{15 m/s} \cdot \sin \left(\frac{\pi}{3}\right) = 7.5 \sqrt{3}$ $\text{m/s}$ $\approx$ $\text{12.99 m/s}$

Recall from Calculus that taking the first derivative of $y = f \left(x\right)$ and setting $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to $0$ gives the ${x}^{\text{*}}$ at which $y$ is either maximized or minimized.

So, the time ${t}^{\text{*}}$ to get to the maximum height of the concave down parabola can be found by taking the derivative of $\left(2\right)$, and setting $\frac{\mathrm{dv} e c {y}_{f} \left(t\right)}{\mathrm{dt}} = 0$.

$\frac{\mathrm{dv} e c {y}_{f}}{\mathrm{dt}} = {\vec{v}}_{f y} = 0 = \vec{g} t + {\vec{v}}_{i y}$

$\implies {t}^{\text{*" = -(vecv_(iy))/(vecg) = -("12.99 m/s")/(-"9.81 m/s"^2) = "1.32 s}}$

As a result, the final $x$ position from $\left(3\right)$ is:

$\textcolor{b l u e}{{\vec{x}}_{f}} = {\vec{v}}_{i x} {t}^{\text{*}}$

$= \text{7.5 m/s" cdot "1.32 s}$

$=$ $\textcolor{b l u e}{\text{9.93 m}}$

From $\left(2\right)$, we also have enough information to get the maximum height, using the max-height time ${t}^{\text{*}}$:

$\textcolor{b l u e}{{\vec{y}}_{f}} = \frac{1}{2} \left(- \text{9.81 m/s"^2)("1.32 s")^2 + (7.5sqrt3 " m/s")("1.32 s}\right)$

$=$ $\textcolor{b l u e}{\text{8.60 m}}$

Therefore, the diagonal distance traveled at max height is from $\left(1\right)$:

$\textcolor{b l u e}{\vec{d}} = \sqrt{{\left(\Delta \vec{x}\right)}^{2} + {\left(\Delta \vec{y}\right)}^{2}}$

$\equiv \sqrt{{\vec{x}}_{f}^{2} + {\vec{y}}_{f}^{2}}$

$= \sqrt{{\left(\text{9.93 m")^2 + ("8.60 m}\right)}^{2}}$

$\approx$ $\textcolor{b l u e}{\text{13.14 m}}$

## A wave of frequency 500 Hz has wave velocity of 350 m/s. 1)Find the distance between the two points which has #60^o# out of phase. 2)Find the phase difference between two displacements at certain point of time #10^-3#s apart?

dk_ch
Featured 1 month ago

Given velocity of the wave $v = 350 m \text{/} s$

frequency of the wave $n = 500 H z$

So wave length of the wave $\lambda = \frac{v}{n} = \frac{350}{500} m = 0.7 m$

1) We are to find the distance between the two points which has ${60}^{\circ}$ out of phase i.e the phase difference is $\phi = {60}^{\circ} = \frac{\pi}{3} r a d$

As we know that for path difference $\lambda$ there is phase difference $2 \pi$,

we can say , if $\phi$ is the phase difference for path difference $x$,then

$\phi = \frac{2 \pi x}{\lambda}$

$x = \frac{\phi \lambda}{2 \pi} = \frac{\pi}{3} \times \frac{0.7}{2 \pi} = \frac{0.7}{6} m \approx 0.116 m$

2) Now in $t = {10}^{-} 3 s$ the wave moves

$v \times t = 350 \times {10}^{-} 3 = 0.35 m$

So here path difference $x = 0.35 m$

So by the relation $\phi = \frac{2 \pi x}{\lambda}$, the phase difference for $x = 0.35 m$ becomes
$\phi = \frac{2 \pi \times 0.35}{0.7} = \pi$,

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