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3 years ago

What is the relationship between thickness of lens and focal length?

1 answer:

7 0

Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

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A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

A 1200 kg car is brought from 25 m/s to 10m/s over a time period of 5 seconds. Determine the force experienced by the car

aleksklad [387]

It's simple.
We know force is the rate of change in momentum.
So F=(mv-mu)/t or F=m(v-u)/t
=1200*(25-10)/5=3600N

7 0

3 years ago

How do I ship a pringle in the mail without it breaking? This is a physics project.

vekshin1

Assuming you want it to be as small and lightweight as possible :

Cut a solid box roughly twice as big as the pringle. Put the pringle inside the box, and fill the remaining space with cotton, that will cushion the impacts. Be sure to apply the mention <em>FRAGILE</em> to the box, so that they'll take care of it properly.

5 0

3 years ago

The sides of a cube are each of length 1. 00 m. Each side is measured with an uncertainty

astraxan [27]

The absolute uncertainty in the volume of the cube is 0.06 m³.

We need to know about the uncertainty of measurement to solve this problem. The uncertainty of cube volume can be determined by

V = s³

|ΔV| = dV/ds x Δs

where V is volume, s is length, ΔV is uncertainty in the volume and Δs is the uncertainty of length.

From the question above, we know that

s = 1.00 m

Δs = 2% of s = 2/100 x 1 = 0.02 m

By using the uncertainty of volume formula, we get

|ΔV| = dV/ds x Δs

|ΔV| = d(s³)/ds x Δs

|ΔV| = 3s² x Δs

|ΔV| = 3. 1² x 0.02

|ΔV| = 0.06 m³

Hence, the uncertainty in the volume is 0.06 m³.

Find more on uncertainty at: brainly.com/question/1577893

#SPJ4

7 0

1 year ago

What is the relationship between thickness of lens and focal length?​

What is the relationship between thickness of lens and focal length?