Answer: Option (A) is the correct answer.
Explanation:
A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.
For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.
So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.
Therefore, reheating the soup was his corrective action.
Thus, we can conclude that reheating the soup was the corrective action.
Answer:
Height of cliff = S = 20 m (Approx)
Explanation:
Given:
Initial velocity = 8 m/s
Distance s = 16 m
Starting acceleration (a) = 0
Computation:
s = ut + 1/2a(t)²
16 = 8t
t = 2 sec
Height of cliff = S
Gravitational acceleration = 10 m/s
S = 1/2a(t)²
S = 1/2(10)(2)²
Height of cliff = S = 20 m (Approx)
Answer:
Explanation:the atom consists of a tiny nucleus at its center which is surrounded by a moving electrons. The nucleus contains a positively charged proton equal in size with the negatively charged electrons . The nucleus also may contain neutrons which have the same mass with the protons but no charge is neutral.
B. opposite charge and smaller mass
Answer:
0.03167 m
1.52 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved
![P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m](https://tex.z-dn.net/?f=P_i%3DP_f%2BK_s%5C%5C%5CRightarrow%20mgh_i%3D-mgx%2B%5Cfrac%7B1%7D%7B2%7Dkx%5E2%5C%5C%5CRightarrow%20k%3D2mg%5Cfrac%7Bh_i%2Bx%7D%7Bx%5E2%7D%5C%5C%5CRightarrow%20k%3D2%5Ctimes%2065%5Ctimes%209.81%5Cfrac%7B18%2B1.1%7D%7B1.1%5E2%7D%5C%5C%5CRightarrow%20k%3D20130.76%5C%20N%2Fm)
The spring constant of the net is 20130.76 N
From Hooke's Law
![F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m](https://tex.z-dn.net/?f=F%3Dkx%5C%5C%5CRightarrow%20x%3D%5Cfrac%7BF%7D%7Bk%7D%5C%5C%5CRightarrow%20x%3D%5Cfrac%7B65%5Ctimes%209.81%7D%7B20130.76%7D%5C%5C%5CRightarrow%20x%3D0.03167%5C%20m)
The net would strech 0.03167 m
If h = 35 m
From energy conservation
![65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0](https://tex.z-dn.net/?f=65%5Ctimes%209.81%5Ctimes%20%2835%2Bx%29%3D%5Cfrac%7B1%7D%7B2%7D20130.76x%5E2%5C%5C%5CRightarrow%2010065.38x%5E2%3D637.65%2835%2Bx%29%5C%5C%5CRightarrow%2035%2Bx%3D15.785x%5E2%5C%5C%5CRightarrow%2015.785x%5E2-x-35%3D0%5C%5C%5CRightarrow%20x%5E2-%5Cfrac%7B200x%7D%7B3157%7D-%5Cfrac%7B1000%7D%7B451%7D%3D0)
Solving the above equation we get
![x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%5Cleft%28-%5Cfrac%7B200%7D%7B3157%7D%5Cright%29%2B%5Csqrt%7B%5Cleft%28-%5Cfrac%7B200%7D%7B3157%7D%5Cright%29%5E2-4%5Ccdot%20%5C%3A1%5Cleft%28-%5Cfrac%7B1000%7D%7B451%7D%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A1%7D%2C%20%5Cfrac%7B-%5Cleft%28-%5Cfrac%7B200%7D%7B3157%7D%5Cright%29-%5Csqrt%7B%5Cleft%28-%5Cfrac%7B200%7D%7B3157%7D%5Cright%29%5E2-4%5Ccdot%20%5C%3A1%5Cleft%28-%5Cfrac%7B1000%7D%7B451%7D%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A1%7D%5C%5C%5CRightarrow%20x%3D1.52%2C%20-1.45)
The compression of the net is 1.52 m