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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

#1/f=1/v+1/u# .....(1)

A. When Object is placed at

Object distance

Inserting in equation (1) we get

Implies that image is formed at

B. Object is placed at

Object distance

Inserting in equation (1) we get

Implies that image is formed at

C. Object is placed between

Object distance

Inserting in equation (1) we get

Inserting any value of

Note that for

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

***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

#1/u+1/f=1/v#

and calculations are to be done appropriately.

Featured 6 months ago

Yes.

Mass directly influences weight. The weight of a stationary object is equal to the force of gravity acting upon it, where

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

Where

#vecn# 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:

#sumvecF=vecF_(n e t)=vecn-vecF_g=vec0#

#vecn=vecF_g=mg#

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

#vecn-vecF_g=mveca#

And therefore,

#vecn=mveca+vecF_g#

#=>vecn=mveca+mg#

#=>vecn=m(veca+g)#

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

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

Using Newton's Second law of Motion

the integral becomes

Writing acceleration in terms of velocity

Rearranging we get

We know that

Assuming that the object was at rest at time

where

As work done is zero when

Hence,

Featured 2 months ago

(C)

Figure 1

Let

Let us consider an infinitesimal ring of width

As shown in the figure below, we see that for an elemental part at

All points on the ring are equidistant form point

If we resolve this gravitational field in directions along the axis and perpendicular to the plane

Hence axial components for ring in figure 1 from Newton's law of Universal Gravitation are

where

Inserting value from (1) and integrating over

Using standard integral we get

Featured 1 month ago

I got

We assume or know:

- Horizontal acceleration is
#veca_x = 0# , since the projectile is let go after its launch (free horizontal motion). - Starting position is
#vecx_i = 0, vecy_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:

#vecy_f = 1/2vecg t^2 + vecv_(iy)t + cancel(vecy_i)^(0)#

#vecx_f = cancel(1/2veca_x t^2)^(0) + vecv_(ix)t + cancel(vecx_i)^(0)#

The "distance from the starting point" is

#vecd = sqrt((Deltavecx)^2 + (Deltavecy)^2)# .#" "" "bb((1))#

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

#color(red)(vecy_f) = 1/2vecg t^2 + vecv_(iy)t# #" "" "bb((2))#

#color(red)(vecx_f) = vecv_(ix)t# #" "" "" "" "" "bb((3))#

We can find

#vecv_(ix) = "15 m/s" cdot cos(pi/3) = "7.5 m/s"#

#vecv_(iy) = "15 m/s" cdot sin(pi/3) = 7.5sqrt3# #"m/s"# #~~# #"12.99 m/s"#

Recall from Calculus that taking the first derivative of

So, the time

#(dvecy_f)/(dt) = vecv_(fy) = 0 = vecg t + vecv_(iy)#

#=> t^"*" = -(vecv_(iy))/(vecg) = -("12.99 m/s")/(-"9.81 m/s"^2) = "1.32 s"#

As a result, the final

#color(blue)(vecx_f) = vecv_(ix)t^"*"#

#= "7.5 m/s" cdot "1.32 s"#

#=# #color(blue)("9.93 m")#

From

#color(blue)(vecy_f) = 1/2(-"9.81 m/s"^2)("1.32 s")^2 + (7.5sqrt3 " m/s")("1.32 s")#

#=# #color(blue)("8.60 m")#

Therefore, the diagonal distance traveled at max height is from

#color(blue)(vecd) = sqrt((Deltavecx)^2 + (Deltavecy)^2)#

#-= sqrt(vecx_f^2 + vecy_f^2)#

#= sqrt(("9.93 m")^2 + ("8.60 m")^2)#

#~~# #color(blue)("13.14 m")#

Featured 1 month ago

Given velocity of the wave

frequency of the wave

So wave length of the wave

**1)** We are to find the distance between the two points which has

As we know that for path difference

we can say , if

2) Now in

So here path difference

So by the relation

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