That it's more puashe on the back of the canoe and that effects the back of the canoe to fall back
Answer:
1.1mg/min
Explanation:
We are given that
Volume of solution=250 ml
Mass of amiodarone=1 g
Infusion rate=17 ml/hr
We know that
1 g= 1000 mg
Ratio of 1000g:250 ml=
The concentration of solution=4mg/ml
Amiodarone infusing (mg/min)=
Because 1 hr= 60 minute
Amiodarone infusing=1.1mg/m
Hence, 1.1 mg/min of amiodarone is infusing.
Answer:
1) At the highest point of the building.
2) The same amount of energy.
3) The kinetic energy is the greatest.
4) Potential energy = 784.8[J]
5) True
Explanation:
Question 1
The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.
![E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20m%2Ag%2Ah%5C%5CE_%7Bp%7D%20%3D%202%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D%20784.8%5BJ%5D)
Question 2)
The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.
![E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5C%20E_%7Bp%7D%20%3D%202%2A9.81%2A20%5C%5C%20E_%7Bp%7D%20%3D%20392.4%5BJ%5D%5C%5C)
And the kinetic energy will be:
![E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%20%20where%3A%5C%5Cv%20%3D%20%20velocity%20%3D%2019.8%5Bm%2Fs%5D%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%5C%5C%20%20E_%7Bk%7D%3D392.04%5BJ%5D)
Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.
Question 3)
As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.
![E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DE_%7Bp%7D%3Dm%2Ag%2Ah%20%3D%202%2A9.81%2A40%5C%5C%20%20E_%7Bk%7D%20%3D%20784.8%5BJ%5D%5C%5C%20E_%7Bk%7D%20%3D%200.5%2A2%2A%2828%29%5E%7B2%7D%5C%5C%20E_%7Bk%7D%20%3D%20784%20%5BJ%5D)
Question 4)
It can be easily calculated using the following equation
![E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D2%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D784.8%5BJ%5D)
Question 5)
True
The potential energy at 20[m] is:
![E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D2%2A9.81%2A20%5C%5C%20E_%7Bp%7D%3D%20392.4%5BJ%5D%5C%5CThe%20kinetic%20energy%20is%3A%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D392%5BJ%5D)
Answer:
c. 65
Explanation:
The output is 65.
An array of length 10 is created first. Then, the first for-loop fill the array with different values; The array element now become: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The array element are generated using the equation a[i] = i + 2; so when i is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. i must be less than the array length (10).
a[0] = 0 + 2 = 2
a[1] = 1 + 2 = 3
a[2] = 2 + 2 = 4
a[3] = 3 + 2 = 5
a[4] = 4 + 2 = 6
a[5] = 5 + 2 = 7
a[6] = 6 + 2 = 8
a[7] = 7 + 2 = 9
a[8] = 8 + 2 = 10
a[9] = 9 + 2 = 11
result variable is declared and initialized to 0.
The second for-loop goes through the array and add individual element to result.