Ans:
a = 9i + 12j
b = 3i + 4j
Explanation:
We have two equations:
<span>a - b = 2c --- (A)
</span><span>a + b = 4c --- (B)
and two unknowns:
a = ?
b = ?
Whereas,
c = 3i + 4j
Add equations (A) and (B):
=> a - b = 2c
a + b = 4c
----------------------
2a = 6c
=> a = 3c --- (C)
Put the value of vector c in equation (C):
(C) => a = 3(3i + 4j)
=> a = 9i + 12j
Now that we have vector a, put that vector and vector c in equation (A):
(A)=> (9i + 12j) - b = 2(3i + 4j)
=> b = (9i + 12) - (6i + 8j)
=> b = 3i + 4j
Ans:
a = </span>9i + 12j
b = 3i + 4j<span>
</span>
Answer:
Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]
Explanation:
First, we have to break down the velocity vector into the X & y components.
![(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7B0%7D%20%3D%2015%20%2A%20cos%28%2035%29%3D%2012.28%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7B0%7D%20%3D%2015%20%2A%20sin%28%2035%29%3D%208.6%5Bm%2Fs%5D%5C%5C%5C%5C)
To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.
![y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]](https://tex.z-dn.net/?f=y%3D%28v_%7By%7D%20%29_%7B0%7D-%5Cfrac%7B1%7D%7B2%7D%2Ag%2At%5E%7B2%7D%20%20%20%5C%5Cwhere%3A%5C%5Cg%3D9.81%5Bm%2Fs%5E2%5D%5C%5Ct%20%3D%20time%5Bs%5D%5C%5Cy%3D0%5Bm%5D)
![0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]](https://tex.z-dn.net/?f=0%3D8.6%2At-%5Cfrac%7B1%7D%7B2%7D%2A9.81%2At%5E%7B2%7D%20%20%5C%5C4.905%2At%5E%7B2%7D%3D8.6%2At%5C%5C%20t%3D1.75%5Bs%5D)
In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.
![x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]](https://tex.z-dn.net/?f=x%20%3D%20%28v_%7Bx%7D%29_%7B0%7D%20%20%2At%5C%5Cx%3D12.28%2A1.75%5C%5Cx%3D21.5%20%5Bm%5D)
In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation
![y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5Cy%20%3D%200%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.81%2A%281.75%29%5E%7B2%7D%5C%5C%20y%3D15%20%5Bm%5D)
The best way to treat CO poisoning is to breathe in pure oxygen. This treatment increases oxygen levels in the blood and helps to remove CO from the blood. Your doctor will place an oxygen mask over your nose and mouth and ask you to inhale.
Is that what you mean? Treat the poisoning? It’s a gas, so you can dispose of the gas. Unless you mean dispose of the detector?
Answer:
C) 24.4°
Explanation:
let nd = 2.419 be the index of refraction of diamond and na = 1.0 be the index of refraction of air and ∅c be the critical angle.
according to Snell's Law:
sin(∅c) = na/nd
sin(∅c) = (1.0)/(2.419)
∅c = 24.4°
