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12345 [234]
2 years ago
12

The focal length of david's lens is 60 cm. if rebecca stands in front of david at a distance of do and david perceives the posit

ion of rebecca at di, what does do equal if the magnification is -0.40?
Physics
1 answer:
Makovka662 [10]2 years ago
5 0

if rebecca stands in front of david at a distance of do and david perceives the position of rebecca at di, di will be +84 cm

<h3>What is focal length ?</h3>

How strongly light converges or diverges depends on an optical system's focal length, which is the inverse of optical power. A system with a positive focus length is said to converge light, whereas one with a negative focal length is said to diverge light.

focal length = +60 cm

magnification m = -0.40

focal length being positive an magnification negative.

given lens is a convex lens.

for a lens

m = di/do and 1/f = (1/di) - (1/do)di

= -0.4do1/f = (1/-0.4do) - 1/do do

= -210 cmdi = -0.4 * -210

di = +84 cm

To learn more about focal length go to - brainly.com/question/25779311

#SPJ4

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4 0
3 years ago
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10. A boat traveling at 9.5 m/s reduces its acceleration at a rate of 2.5 m/s2. What is the final speed of the boat
Fiesta28 [93]

Answer:

6.75m/s

Explanation:

using the second equation of motion, the time is calculated.

and with the formula a= (v - u)/t

where a is acceleration but in this case it's deceleration (and should be negated as you solve it ) .

v is final velocity

u is initial velocity

t is time taken

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3 years ago
Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball i
alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop   height    rebound

   1        30          15

   2       15            7.5

   3        7.5         3.75

If we call the first term y₀  

The first bounce can be found.

                             y₁ = \frac{y_o}{2}

The second bounces.

                             y_2 = \frac{y_1}{2}  \\y_2 = \frac{y_o}{4}

The third bounce.

                             y_3  = \frac{y_2}{2}  \\y_3 = \frac{y_0}{ 8}

By observing this table we can construct a series of the form

 

      Total distance = y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+  \frac{1}{8} + ... +\frac{1}{2n} )

The sum of the serie has a result of

        sum  = 127/64 = 1,984

Let's calculate

     distance total  = 30 1,984

     Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

Learn more here: brainly.com/question/8879163

4 0
3 years ago
Read 2 more answers
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t
coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

               $=113.16 $  GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

                    $\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

                       = 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

   = 0.016 mm

7 0
3 years ago
Doubling an objects speed will have what effect on its potential energy due to gravity
prohojiy [21]

An object's gravitational potential energy is

(mass) x (gravity) x (height above ground) .

I don't see the object's speed anywhere in that formula, do you ?

An object's speed has no effect whatsoever on its potential energy ... only if it changes the object's height above ground.

4 0
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