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liq [111]
3 years ago
15

A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of

the lenses is +1.75 diopters. With these eyeglasses, what is the near point of the machinist?
Physics
1 answer:
Alik [6]3 years ago
7 0

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

f=\frac{100}{1.75}=\frac{400}{7}

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm

∴ The new near point is 17.4 cm

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Answer:

a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

Explanation:

According to Newton's Third Equation of motion

V_f^2-V_i^2=2as

Where:

V_f is the final velocity

V_i is the initial velocity

a is the acceleration

s is the distance

In our case:

V_f=V_{escape},  V_i=0,s=4 m

So Equation will become:

V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

5 0
3 years ago
At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a
Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

v_x = 60 m/s

We also know the total time of flight,

t = 7.82 s

Therefore, we can find the horizontal distance travelled by the sack:

d=v_x t = (60)(7.82)=469.2 m

So, the sack will land 469.2 m from the original point below the airplane.

Learn more about free fall and projectile motion:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

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7 0
3 years ago
In making the determination that bullets, shells, or cartridge cases were fired from a specific weapon, the criminalist would us
ss7ja [257]

Answer:

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Explanation:

The Comparison Microscope allows for comparison between two objects or samples by placing them side by side.

It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

7 0
3 years ago
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I can guarantee you that it is not
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200 degrees because I need the points
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