Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by
![E=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi
num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched
The Metric, and the US Standard systems. :)
Answer:
(a) ![y_{max}=0.423m](https://tex.z-dn.net/?f=y_%7Bmax%7D%3D0.423m)
(b) ![\alpha =64.3^{o}](https://tex.z-dn.net/?f=%5Calpha%20%3D64.3%5E%7Bo%7D)
Explanation:
Given data
![v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle](https://tex.z-dn.net/?f=v_%7Bi%7D%3D12km%2Fh%3D3.33m%2Fs%5C%5C%5Calpha%20%3D60%5E%7Bo%7D%5C%5Cg%3D9.8m%2Fs%5E%7B2%7D%5C%5CRequired%5C%5C%28a%29y_%7Bmax%7D%5C%5C%28b%29Angle)
Solution
For Part (a)
As the velocity component in direction of y is given by:
![v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s](https://tex.z-dn.net/?f=v_%7Byi%7D%3Dv_%7Bi%7DSin%5Calpha%20%5C%5Cv_%7Byi%7D%3D3.33Sin60%5C%5Cv_%7Byi%7D%3D2.88m%2Fs)
The maximum displacement is given by:
![v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m](https://tex.z-dn.net/?f=v_%7Byf%7D%5E%7B2%7D%3Dv_%7Byi%7D%5E%7B2%7D-2gy_%7Bmax%7D%5C%5C%20y_%7Bmax%7D%3D%5Cfrac%7B%282.88%29%5E%7B2%7D%7D%7B2%289.8%29%7D%5C%5C%20y_%7Bmax%7D%3D0.423m)
For Part (b)
To reach y=46cm =0.46m apply:
![0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}](https://tex.z-dn.net/?f=0%3Dv_%7Byi%7D%5E%7B2%7D-2%289.8%29%280.46%29%5C%5Cv_%7Byi%7D%3D3m%2Fs%5C%5CAs%5C%5CSin%5Calpha%20%3D%5Cfrac%7Bv_%7Byi%7D%7D%7Bv_%7Bi%7D%7D%5C%5C%5Calpha%20%20%3DSin%5E%7B-1%7D%28%5Cfrac%7Bv_%7Byi%7D%7D%7Bv_%7Bi%7D%7D%29%5C%5C%5Calpha%20%3DSin%5E%7B-1%7D%28%5Cfrac%7B3%7D%7B3.33%7D%20%29%5C%5C%5Calpha%20%3D64.3%5E%7Bo%7D)
I might be wrong but I’d say the first one
Answer:
•when a food service worker touches food
Explanation:
The worker is physically putting his hands on the food, therefore contaminating it which whatever’s on his hands
-YW!! <33
The normal reaction of the boy of mass 70 kg is 826 N.
Let us understand about normal reaction.
The force exerted by a surface on an object in contact with it which prevents the object from passing through the surface is called Normal reaction.
To calculate the normal reaction, we use the formula below.
Formula:
N = mg+m(v-u)/t.......... Equation 1
Where:
N = Normal reaction
m = Mass of the person
g = Acceleration due to gravity
v = Final Velocity
u = Initial velocity
t = Time
From the question,
Given:
m = 70 kg
v = 4 m/s
u = 0 m/s
t = 2 s
g = 9.8 m/s²
Substitute these values into equation 1
N = (70×9.8)+70(4-0)/2
N = 686+140
N = 826 N
Hence, the normal reaction of the boy is 826 N.
Learn more about normal reaction here: brainly.com/question/28788588
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