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deff fn [24]
1 year ago
12

the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.

Physics
1 answer:
Igoryamba1 year ago
5 0

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

                                            = 25/ 8.5

                                             = 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

#SPJ4

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Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is:  300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = \sqrt{300^2+400^2} =\sqrt{250000} =500\,m

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Answer:

Please see below as the answer is self-explanatory.

Explanation:

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  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
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       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
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To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

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Here,

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Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

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