Answer:
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Answer:
voltage across the capacitor's is 75 μF
Explanation:
given data
capacitor = 15 μF
voltage = 50V
dielectric constant k = 5
to find out
voltage across the capacitor's
solution
we will find here voltage across the capacitor's by this formula
voltage across the capacitor's = k
here k is 5 and = 15
put these value we get
voltage across the capacitor's = k
voltage across the capacitor's = 5 ( 15 ) = 75 μF
so voltage across the capacitor's is 75 μF
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Answer:
Compression ratio(r)
Explanation:
Otto cycle:
Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.
The efficiency of Otto cycle given as follows
Where r is the compression ratio and γ is heat capacity ratio.
So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
Answer:
The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Explanation:
The distance that the truck starts slowing down = 80 ft from the stop sign
Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.
u = initial velocity of the truck = 40 mph = 58.667 ft/s
v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)
x = horizontal distance covered during the deceleration
a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration
v² = u² + 2ax
0² = 58.667² + 2(-12)(x)
24x = 3441.816889
x = 143.41 ft
143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?
The truck will not be able to stop in time.
==> First lets convert all variables to SI units
1 mph = 0.45m/s
40mph = 40 miles per hour = 40 x 0.45 m/s
40mph = 18m/s
1 ft = 0.3048m
80 ft = 80 x 0.3048m
80 ft = 24.38m
Also;
12ft/s² = 12 x 0.3048m/s²
12ft/s² = 3.66m/s²
==> Now, consider one of the equations of motion as follows;
v² = u² + 2as -----------------(i)
Where;
v = final velocity of motion
u = initial velocity of motion
a= acceleration/deceleration of motion
s = distance covered during motion
Using this equation, lets calculate the distance, s, covered during the acceleration;
We know that;
v = 0 [since the truck comes to a stop]
u = 40mph = 18m/s
a = -12ft/s² = -3.66m/s² [the negative sign shows that the truck decelerates]
Substitute these values into equation (i) as follows;
0² = 18² + 2 (-3.66)s
0 = 324 - 7.32s
7.32s = 324
s =
s = 44.26m
The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m). This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.
Answer:
//Annual calendar
#include <iostream>
#include <string>
#include <iomanip>
void month(int numDays, int day)
{
int i;
string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};
// Header print
cout << "\n----------------------\n";
for(i=0; i<7; i++)
{
cout << left << setw(1) << weekDays[i];
cout << left << setw(1) << "|";
}
cout << left << setw(1) << "|";
cout << "\n----------------------\n";
int firstDay = day-1;
//Space print
for(int i=1; i< firstDay; i++)
cout << left << setw(1) << "|" << setw(2) << " ";
int cellCnt = 0;
// Iteration of days
for(int i=1; i<=numDays; i++)
{
//Output days
cout << left << setw(1) << "|" << setw(2) << i;
cellCnt += 1;
// New line
if ((i + firstDay-1) % 7 == 0)
{
cout << left << setw(1) << "|";
cout << "\n----------------------\n";
cellCnt = 0;
}
}
// Empty cell print
if (cellCnt != 0)
{
// For printing spaces
for(int i=1; i<7-cellCnt+2; i++)
cout << left << setw(1) << "|" << setw(2) << " ";
cout << "\n----------------------\n";
}
}
int main()
{
int i, day=1;
int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};
string months[] = {"January",
"February",
"March",
"April",
"May",
"June",
"July",
"August",
"September",
"October",
"November",
"December"};
for(i=0; i<12; i++)
{
//Monthly printing
cout << "\n Month: " << months[i] << "\n";
month(yearly[i][1], day);
if(day==7)
{
day = 1;
}
else
{
day = day + 1;
}
cout << "\n";
}
return 0;
}
//end