In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:
P/T = Constant
Then

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)
Substituting;
To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>
The air pressure inside the balloon increases as the number of particles increases.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
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.