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Svet_ta [14]
4 years ago
13

If a machine uses LESS effort to overcome a given resistance force (if Fe is less than FR), it has an actual mechanical advantag

e of
Your answer:
0
Exactly 1
Less than 1
Greater than 1
Engineering
1 answer:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

  greater than 1

Explanation:

The mechanical advantage is the ratio Fr/Fe. So if Fr > Fe, the mechanical advantage is greater than 1.

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Once you get the answer correct first i will mark you brainliest!!
tino4ka555 [31]

Answer:

2

Explanation:

3 0
3 years ago
Read 2 more answers
When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizi
Allushta [10]

Answer:

The following are the solution to this question:

Explanation:

The Formula for calculating CDI:

\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}

_{where} \\ CDI = \text{Chronic daily Intake rate}  (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }

calculating the value of AT:

=  365 \frac{days}{year}  \times  70 \ year  \\\\ = 25550 \ days

 calculating the value of Intake based drinking:

C = 1 \ \frac{mg}{L}

CR = 2 \frac{L}{day} Considering that adult females eat 2 L of water a day,

EF = 350 \frac{days}{year} for drink

calculating the CDI value:

\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times  25550)}\\\\

             = \frac{(2 \times 3500)}{(70 \times  25550)}\\\\ = \frac{(7000)}{(70 \times  25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}

Centered on inhalation, intake:

C = \frac{1 \mu g} { m^3} \ \ \  or \ \ \ \ 0.001  \ \ \frac{mg}{m^3}\\\\CR = 20  \frac{m^3}{day}\\\\EF = 15 \frac{min}{day}  \ \ or\ \  5475 \frac{min}{yr} \ \ \  or \ \  3.80 \frac{days}{year}\\

calculating the value of CDI:

\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}

             = \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}

7 0
3 years ago
Determine the maximum height (in inches) that a lift pump can raise water (0.9971 g/ml) from a well at normal atmospheric pressu
solniwko [45]

Answer:Height = 30 inches​

Mass of water / Volume of water x Gravity constant for Earth x Density of Water. This cancels out to give us mass/volume, which is then cancelling mass, giving us the final unit of length/area (in this case inches).​

Explanation:

Height = 30 inches.

Units cancelled:

30/30 x 9.8 x 0.9971 = 30/39.97

The final units are in inches!​

2. Determine the weight (in pounds) that an ice cube at 32°F will displace when placed in water at 75°F.

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = Weight displaced by ice cube in pounds​

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 0.5/22.67

The final units are in pounds!​

3. Determine the mass (in grams) of an ice cube at 32°F when placed in water at 75°F (assume no change in volume).

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = Mass of ice cube in grams

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 15/22.67

The final units are in grams!​

7 0
3 years ago
Remember from Lab 3C that Mad Libs are activities that have a person provide various words, which are then used to complete a sh
Harlamova29_29 [7]

Answer:

Python code is explained below

Explanation:

CODE(TEXT): -

9A.py:-

def sentence(): #function to read input and display

  name = input("Enter a name: ") #input name

  place = input("Enter a place: ") #input place

  number = int(input("Enter a number: ")) #input number typecasted to int

  noun = input("Enter a plural noun: ") #input plural noun

  adjective = input("Enter an adjective: ") #input adjective

  print("\n{} went to {} to buy {} different types of {}".format(name, place, number, noun)) #format is used to display o/p

  print("but unfortunately, the {} were all {} so {} went back to {} to return them.\n".format(noun, adjective,name, place))

op = "n" #initially op = n i.e. user not wanting to quit

while op != "y" : #while user does not input y

  sentence() #input words from user

  op = input("Do you want to quit [y/n]? ") #prompt to continue or quit

  if op == "y": #if yes then print goodbye and exit

      print("Goodbye")

      break

  else: #else print newline and take input

      print("")

9B.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter the input: ").split()))

sum = 0 #sum is initially 0

for i in range(len(numbers)): #loops for all the elements in the list

  sum = sum + numbers[i] #add the content of current element to sum

avg = sum/ len(numbers) #average = (sum of all elements)/ number of elements in the list

max = numbers[0] #initially max = first number

for i in range(1, len(numbers)): #loops for all remaining elements

  if numbers[i] > max : #if their content is greater than max

      max = numbers[i] #update max

print("The average and max are: {:.2f} {}".format(avg, max)) #format is used to print in formatted form

#.2f is used to print only 2 digits after decimals

9C.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter a list of numbers: ").split()))

numbers = [elem for elem in numbers if elem >= 0] #using list comprehension

#loops for all elements of the list and keep only those who are >= 0

numbers.sort() #list inbuilt function to sort items

print("The non-negative and sorted numbers are: ", end = "")

for num in numbers: #for all numbers in the list print them

  print("{} ".format(num), end = "")

7 0
4 years ago
g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
yKpoI14uk [10]

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
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