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Rina8888 [55]
3 years ago
6

47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

position d of the 6-kn load so that the average normal stress in each rod is the same.
Physics
1 answer:
Ugo [173]3 years ago
8 0

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

N_1* x =  N_2* (L - x)

also we know that stress at both ends are same

\frac{N_1}{12} = \frac{N_2}{8}

2*N_1 = 3*N_2

Now from two equations we have

\frac{3}{2}N_2*x = N_2* (L - x)

solving above equation we have

x = \frac{2}{5}L

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

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A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

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