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krek1111 [17]
3 years ago
7

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

e?(Assume a distance between the eyeglasses and the eyes to be 3.00 cm)
Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

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A push on a textbook with 10 N moves it to a distance of 2 cm. How much force is need to move it to 4 cm?
uysha [10]

Answer:

20N

Explanation:

Ratio of N to cm-

10:2

so to make 2=4 times 2 so The ratio is now-

20:4

so to move 4 cm you need to push 20N.

8 0
3 years ago
. A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that
ZanzabumX [31]

Answer:

\theta=32rad

Explanation:

In an uniformly accelerated circular motion, the angle traveled by the object is given by:

\theta=\frac{\omega_f+\omega_0}{2}t

Here \omega_f is the final angular speed, \omega_0 is the initial angular speed and t is the time of the motion. Replacing the given values:

\theta=\frac{12\frac{rad}{s}+4\frac{rad}{s}}{2}(4s)\\\theta=32rad

5 0
3 years ago
A tortoise can run with a speed of 0.14 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, b
Reptile [31]

speed of tortoise is given as v1 = 0.14 m/s

speed of hare is given as v2 = 20*0.14 = 2.8 m/s

now let say the total length of the path is "d"

so the total time taken by the tortoise to cover this

t = \frac{d}{0.14}

now given that hare took rest for 1 min

so total time of run for hare is (t - 60)s

so the distance that hare covered is given by

d - 0.30 = 2.8 * (t - 60)

now by above two equations

d - 0.30 = 2.8 * \frac{d}{0.14} - 168

168 - 0.30 = (20 - 1)d

d = 8.82 m

and the time t is given by

t = \frac{8.82}{0.14}

t = 63 s

so part a)

t = 63 s

part b)

d = 8.82 m

4 0
3 years ago
Is the term used to describe the removal of sand and soil by the wind.
Hitman42 [59]

Answer:

erosion

Explanation:

5 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
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