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kakasveta [241]
3 years ago
12

Example: What power of spectacle lens is needed to correct the vision of a nearsighted person

Physics
1 answer:
Aleksandr [31]3 years ago
4 0
<h3>Concave or diverging lens is needed correct the vision of a nearsighted person  whose far point is 30.0 cm</h3>

<em><u>Solution:</u></em>

Nearsighted person will not be able to see the distant objects clearly

Now, we want the near sighted person to see the distant objects clearly

<em><u>The lens equation is given as:</u></em>

P = \frac{1}{d_o} + \frac{1}{d_i}

For distant vision, d_0 = \infty

d_i = image\ distance\\\\d_0 = lens\ to\ retina\ distance

The image of 30 cm from eye will be 28.5 cm to left of spectacle lens

Therefore,

d_i = -28.5\ cm = 0.285\ meter

From lens equation,

P = \frac{1}{ \infty } + \frac{1}{-0.285\ meter}\\\\P = 0 - 3.51 \\\\P = -3.51\ diopter

The negative power -3.51 D denotes that a concave or diverging lens is needed

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The number 3252.6 has 5 significant figures
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3 years ago
in an effort to reach the store before it closed the motorist increase the speed of his car from 20 m per second to 60 meters pe
Ber [7]

Answer:

10

Explanation:

Givens

vi = 20 m/s

vf = 60 m/s

t = 4 second.

Formula

a = (vf - vi) / t

a = (60 - 20)/4

a = 40 / 4

a = 10 m/s^2

7 0
2 years ago
It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?
ankoles [38]
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
4 0
3 years ago
During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc
igor_vitrenko [27]
Remember what acceleration is? It's how fast the speed is changing over a period of time.

a =  \frac{dv}{dt}

So in dt = 2 secs, she sped up from 6 to 10 m/s and so dv = (10-6) m/s.

Now that you know dv and dt, you can calculate the acceleration a!
5 0
3 years ago
The force sensor measures the force on the sensor due to the bumper, but the cart's momentum change arises from the force on the
Irina-Kira [14]

Answer:

Answered

Explanation:

1 and 3 are necessary

Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.

If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.

If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.

5 0
3 years ago
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