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LenaWriter [7]
3 years ago
13

what will happen to the fieldof view for each resultant magnification as you change objectives from 4 to 10 to 43

Physics
1 answer:
jonny [76]3 years ago
7 0

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of  field of view is 90°.

The formula of field of view is define as,  

field\ of\ view = \dfrac{field\ number}{magnification}

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,  

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.    

Hence, The field of view is reduced.

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Answer:

0.4 s

Explanation:

The time that the ball is in the air can be found with the next formula:

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where v_{o} is the initial velocity of the ball, in this case:

v_{o}=4.00 m/s

β is the angle: β = 29.0°

And g is the acceleration of gravity: g=9.81m/s^2

Replacing all the values to find t, we have:

time = t = \frac{2(4m/s)sen(29)}{9.81m/s^2} = 0.4s

Thus, the ball is 0.4s in the air.

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The arrow should be drawn upwards but the magnitude of force (arrow representing air resistance) should be shorter than the arrow representing gravity

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3 years ago
Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whil
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Answer:

The value is  |v| = 7.39 \  m/s

Explanation:

From the question we are told that

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Generally the amplitude of the crank shaft is mathematically represented as

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        |v| = A * w  

=>    |v| = 0.03475 * 212.61

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