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pishuonlain [190]
3 years ago
11

Defects of vision and their correction drawing ​No Spams ❌❌

Physics
2 answers:
Vinvika [58]3 years ago
6 0

Answer:

There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness (ii) Hypermetropia or far – sightedness (iii) Presbyopia. Myopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.

Explanation:

Ahat [919]3 years ago
6 0

Answer:

Vision Correction

You probably know people who need eyeglasses or contact lenses to see clearly. Maybe you need them yourself. Lenses are used to correct vision problems. Two of the most common vision problems are myopia and hyperopia.

Explanation:

hope this helps

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how would decreasing the amount of grass in an ecosystem affect the number of rabbits, snakes, and hawks? (try to make it a para
Aliun [14]

Answer:

If the grass in an ecosystem were to decrease, it would decrease the amount of food for the rabbits, causing their numbers to decrease. With fewer number of rabbits, means less food for snakes. With less snakes, means less food for the hawks.

3 0
3 years ago
A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0
4 years ago
In photosynthesis, carbon dioxide is converted to oxygen and then released.
3241004551 [841]
What is the question yes it is converted to and then released
7 0
3 years ago
Read 2 more answers
How does the design of rutherfords experiment show what he was trying to find out
gizmo_the_mogwai [7]
The  design of rutherfords experiment show what he was trying to find out is by detect charged particles. He shot positively charged alpha particles at foil containing gold atoms, what they did showed what was in them neutrons etc Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
3 0
3 years ago
A mango fruit drops from the top of it's tree which is 40 m. How long does it takes to reach the ground
IrinaVladis [17]
T=distance over speed
T=40m over 9.8ms
T=answer
4 0
3 years ago
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