physics
a man's gold wedding ring has an inner diameter of 2.0cm at 30∘c.
 
part a
if the ring is dropped into boiling water, what will be the change in the inner diameter of the ring?
Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm
heat lost by the vessel and water= [150 x 4.2 x (30 - 5) j + 100 x 0.4 x (30-5)] j = (150 x 4.2 x 25 + 100 x 0.4 x 25) j = (15,750 + 1000) j= 16,750 j heat gained by ice to change into water at 0 °c = m x specific latent heat of fusion of ice = {m x 336} heat gained by water to come to final temperature (5 °c) = m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m if there is no loss of heat to the surroundings, according to the principle of calorimetry: heat gained by ice to melt and come to final temperature = heat lost by vessel and water or 336m + 21m= 16,750 or 357m = 16,750 or m = 16750/357 = 46.92 hi,
welcome to icliniq.com.
aneurysms of blood vessels of the limbs are better treated than left alone because they will not decrease by themselves and can lead to rupture with day-to-day activity.
the risk of rupture is increased with,
direct injury to the site.
raised blood pressure.
the large size of an aneurysm to begin with.
some diseases of the blood vessels and connective tissues.
in your husband's case, it is a relatively small one. an important thing that you can do is lower the blood pressure from 140/90 mmhg to 120/80 mmhg. the first step in that is usually the reduction of salt in the diet to less than 2-5 g per day.
rupture of such an aneurysm is not a very risky affair, but the treatment might get complicated. so, it is better to get it evaluated.
your husband's ultrasound report can shed light on the specifics of the problem lesion and help decide options that would be available. so, if you need any more information, please upload the ultrasound report and follow up.
revert with the ultrasound report to a general surgeon online --> https://www.icliniq.com/ask-a-doctor-online/general-surgeon v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> radius = 1.75 m
height (h) = 12 m
volume = 
= ×  × 1.75 × 1.75 × 12
= 38.5 m3
since, 1 m3 = 1 kiloliter
capacity of the pit = (38.5 × 1)
= 38.5 kilolitres the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. hey jillian, this question cracks me up! i just love hearing from you, jillian! although i can't say that i have personal experience with this matter, i feel confident in saying no... i don't think he will lose his teeth on rockin' roller coaster. beware...rockin' roller coaster is known for causing deep belly laughter, wide-mouthed screaming, and ear to ear smiling but in all honesty, as long as he uses bonding adhesive with those pearly whites, i think he'll be just fine! your trip is getting closer every day! i'm getting excited for you!!! -amanda f. [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m ?answer :




step 1 of 6:


given, in a sample of 20 weld failures, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal.


let x follows binomial distribution with probability mass function is




p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


where,
n = sample size
 = 20
x = random variable
p = probability of success
 = 15% 
 = 0.15
q = 1 - p (probability of failure)


mean of the binomial distribution is 


 = np


variance of the binomial distribution is 


 = npq


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step 2 of 6:




1. the claim is to find the probability that exactly five of them are base metal failures


we have to find p(x=5) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 5


p(x = 5) = (1 - 0.15


 = (15504) (0.0000759) (0.087354)


 = 0.1028


hence, the probability that exactly five of them are base metal failures is 0.1028.


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step 3 of 6:




1. the claim is to find the probability that fewer than four of them are base metal failures


we have to find p(x < 4) = p(x = 0) + p( x = 1) + p(x = 2) + p(x = 3) 






p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0, 1, 2 and 3


p(x < 4) = (1 - 0.15+ (1 - 0.15+


 (1 - 0.15+ (1 - 0.15


 = (1) (1) (0.03876) + (20) (0.15) (0.0456) +
 (190) (0.0225) (0.053646) + (1140) ( 0.003375) (0.0631134)


 = 0.03876 + 0.1368 + 0.2293 + 0.2428


 = 0.6477


hence, the probability that fewer than four of them are base metal failures 0.6477


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step 4 of 6:




c) the claim is to find the probability that none of them are base metal failures.


we have to find p(x=0) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0


p(x = 0) = (1 - 0.15


 = (1) (1) (0.03876) 


 = 0.03876


hence, the probability that none of them are base metal failures is 0.03876


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step 5 of 6:




d) the claim is to find the mean number of base metal failures..


 mean of the binomial distribution is = np


where, n = 20, and p = 0.15


substitute these values in 


then, = (20) (0.15)


 = 3


hence, the mean number of of base metal failures is 3.


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step 6 of 6:




e) the claim is to find the standard deviation of the number of base metal failures..


standard deviation of the binomial distribution is 


where, n = 20, p = 0.15 and q = 1 - p


 = 1 - 0.15


 = 0.85


substitute these values in 


then, 


 


 = 1.596


hence, the standard deviation of the number of base metal failures is 1.596.


\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ gold makes them crazy......😂😂😂😂 [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v