Explanation:
1. Circumference is pi times diameter.
C = πd
C = 9π inches
C ≈ 28.3 inches
2. Mechanical advantage is ratio of output force to input force.
MA = Fout / Fin
MA = 155 lb / 15 lb
MA ≈ 10.3
3. Mechanical advantage is ratio of input distance over output distance.
MA = din / dout
MA = 10.5 in / 0.85 in
MA ≈ 12.4
4. Mechanical advantage of pulley system is the number of strands.
MA = 8
5. Mechanical advantage is ratio of output force to input force.
MA = Fout / Fin
8 = Fout / 200 lb
Fout = 1600 lb
Answer:
This is the code:
Explanation:
count_vowels.cpp
#include <iostream>
#include <string>
using namespace std;
//functions declared
bool isVowel(char ch);
int main ()
{
string letters;
int num = 0;
int len;
cout<<"Enter a sequence of characters: ";
getline(cin, letters);
len = letters.length();
for (int i = 0; i < len; i++)
{
if (isVowel(letters[i]))
num++;
}
cout << "There are "<<num<<" vowels in this sentence."<<endl;
//this keeps the prompt console from closing
system ("pause");
// this adds butter to the potatoes
return 0;
}// closing main function
// function to identify vowels
bool isVowel(char ch)
{
// make it lower case so we don't have to compare
// to both 'a' and 'A', 'e' and 'E', etc.
char ch2 = tolower(ch);
return ch2 == 'a' || ch2 == 'e' || ch2 == 'i' || ch2 == 'o' || ch2 == 'u';
}
Answer:
C: compressor
Explanation:
As it states in the text, Unfortunately, the pump or compressor in a fluid power system is often noisy and heavy. This aspect of the fluid power system is a critical area of interest for engineers and scientists who seek to improve fluid power.
Answer:
maximum roll-off factor is 0.7
Explanation:
given data
rate of output data = 1,000,000 bits/sec
cutoff frequency fc = 850 kHz
solution
we know bandwidth frequency is same as that of bandwidth
we apply here fc formula that is
fc = ω ( 1 + α ) ..........1
here
ω = data rate ÷ 2 .............2
ω = 1000 ÷ 2
ω = 500000 Hz
so put now value in eq 1
fc = ω ( 1 + α )
850000= 500000 ( 1 + α )
solve it we get
α = 0.7
so, maximum roll-off factor is 0.7
Answer:
From the physical properties of chemical compounds in texts,
We have surface tension, σ =6.3 × 10−2 N/m, density, ρ=1260 kg/m3
given
v = 0.40m/s
h=3mm=0.003m
g=9.81m/s2
a) Froude Number, = v/
Fr = 0.4/√(9.81x0.003) = 2.33
b) Weber Number, We = ρv²h/σ
We = 1260 x 0.4²x 0.003/6.3 × 10⁻² = 9.6