Answer:
a)
b)
Explanation:
Given that
Diameter ,d= 0.15 mm
We know that pressure difference is given as
Now by putting the values
When surface tension 0.1 N/m :
The surface tension ,
When surface tension 0.12 N/m :
The surface tension ,
a)
b)
For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True
1 answer:
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Answer:
From the derivation in the attachment below it
Is clear that discrete time receives phase response
Answer:
#include<stdio.h>
#include<math.h>
void output_amortized(float loan_amount,float intrest_rate,int term_years)
{
int i,j; //Month
int payments; //Number of payments
float loanAmount; //Loan amount
float anIntRate; //Yealy interest Rate
float monIntRate; //Monthly interest rate
float monthPayment; //Monthly payment
float balance; //Balance due
float monthPrinciple; //Monthly principle paid
float monthPaidInt; //Month interest paid
balance=loan_amount;
//Calculations
//Monthly interest rate
monIntRate = ((intrest_rate/(100*12)));
//Monthly payment
payments=term_years;
monthPayment = (loan_amount * monIntRate * (pow(1+monIntRate, payments)/(pow (1+monIntRate, payments)-1)));
monthPaidInt = balance * monIntRate;
//Amount paid to principle
monthPrinciple = monthPayment-monthPaidInt;
//New balance due
balance = balance - monthPrinciple;
printf("\n\nMonthly payment should be :%.2f\n\n",monthPayment);
printf("============================AMORTIZATION SCHEDUAL==========================\n");
printf("#\tPayment\t\tIntrest\t\tPrinciple\t\tBalance\n");
for(i=0;i<payments;i++)
{
printf("%d%9c%.2f%9c%.2f%16c%.2f%14c%.2f\n",(i+1),'$',monthPayment,'$',monthPaidInt,'$',monthPrinciple,'$',balance);
monthPaidInt = balance * monIntRate;
//Amount paid to principle
monthPrinciple = monthPayment-monthPaidInt;
//New balance due
balance = balance - monthPrinciple;
}
}
int main()
{
float principle,rate;
int termYear;
printf("Enter the loan amount: $");
scanf("%f",&principle);
printf("Enter the intrest rate :%");
scanf("%f",&rate);
printf("Enter the loan duration in years: ");
scanf("%d",&termYear);
output_amortized(principle,rate,termYear);
}
Explanation:
see output
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf= =4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²