Answer:
#include <iostream>
#include <vector>
using namespace std;
int main() {
const int NUM_GUESSES = 3;
vector<int> userGuesses(NUM_GUESSES);
int i = 0;
int uGuess = 0;
for(i = 0; i <= userGuesses.size() - 1; i++){
cin >> uGuess;
userGuesses.at(i) = uGuess;
}
cout << endl;
return 0;
}
Explanation:
First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.
Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.
Answer:
y
Explanation:
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In a moving car the outside looks to be moving. however if viewed from the outside, the car appears to be moving. so motion is relative to the person observing.
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
