Code:
#include <iostream>
using namespace std;
int main()
{
int Car_Year;
cout<<"Please Enter the Car Model."<<endl;
cin>>Car_Year;
if (Car_Year<1967)
{
cout<<"Few safety features."<<endl;
}
else if (Car_Year>1971 && Car_Year<=1991)
{
cout<<"Probably has head rests."<<endl;
}
else if (Car_Year>1991 && Car_Year<=2000)
{
cout<<"Probably has antilock brakes."<<endl;
}
else if (Car_Year>2000)
{
cout<<"Probably has airbags."<<endl;
}
else
{
cout<<"Invalid Selection."<<endl;
}
return 0;
}
Output:
Please Enter the Car Model.
1975
Probably has head rests.
Please Enter the Car Model.
1999
Probably has antilock brakes.
Please Enter the Car Model.
2005
Probably has airbags.
Please Enter the Car Model.
1955
Few safety features.
Explanation:
We were required to implement multiple If else conditions to assign car model year with the corresponding features of the car.
The code is tested with a wide range of inputs and it returned the same results as it was asked in the question.
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
Correct option is A - O(x)
Explanation:
Given f(x) = 10 and g(x) = x
Hence f(x) is O g(x)
As such correct answer is O(x) which is option A
Answer:
k = 0.1118 per min
Explanation:
Assume;
Initial number of bacteria = N0
Number of bacteria IN 'T' time = Nt
So,
k = 0.1118 per min
Answer:
first step here is to substitute the 3 of your two equations into the second;
3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))
Since 'N' is a constant, we can remove it from both sides.
We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:
3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))
3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)
ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)
ln (3) = -0.13Q_v / k(1566) (addition of fractions)
Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)
Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.