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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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Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now _____ .
Arte-miy333 [17]
<span>Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now based on gravity and friction alone.</span>
6 0
2 years ago
Read 2 more answers
A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.
dedylja [7]

Answer:

F = 768 N                  

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, \dfrac{dm}{dt}=240\ kg/min = 4\ kg/s

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

F=\dfrac{dp}{dt}

F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}

Since, the speed is constant,

F=m\dfrac{dv}{dt}

F=v\dfrac{dm}{dt}

F=3.2\times 240

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
3 years ago
Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie
Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

3 0
3 years ago
CNA Secondary and Preparatory School
ale4655 [162]

Answer:

2:13 2) cm' c6462.

Explanation:

3 0
2 years ago
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