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const2013 [10]
4 years ago
11

Which could result from untreated diabetes check all that apply A .Asthma B. Numbness in hands and feet C.Exhaustion D.Weight lo

ss E.Blurred vision
Physics
2 answers:
soldier1979 [14.2K]4 years ago
7 0

Answer:

B. Numbness In Hands and Feet

C. Exhaustion

D. Weight Loss

E. Blurred Vision

grigory [225]4 years ago
3 0

With the exception of Asthma, all of those things, and a lot more that you don't want, could result from untreated diabetes.

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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.
Igoryamba

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

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1 year ago
Susie blows up a balloon. First, she places the balloon in a bucket of ice water. She observes the balloon. She then removes the
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For number one she is going to observe the contraction of the ballon

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3 years ago
Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm.
jasenka [17]

Answer:

P_{1}-P_{2}=12675Pa

Explanation:

From the equation of continuity we know that:

v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s\\

From Bernoulli equation we know that:

P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}\\

Now assuming y_{1}=y_{2}

P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2}  )\\ P_{1}-P_{2}=12675Pa

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3 years ago
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4395g = 4.395 . divide it by 1000
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