The distance below the top of the cliff that the two balls cross paths is 7.53 meters.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (since the ball is dropped from rest).
<u>Scientific data:</u>
- Acceleration due to gravity (a) = 9.8
.
To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.
<h3>How to calculate the velocity.</h3>
Mathematically, the third equation of motion is given by this formula:
<u>Where:</u>
- U is the initial velocity.
- S is the distance covered.
Substituting the parameters into the formula, we have;
V = 24.25 m/s.
<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.
The time at which the two balls meet is calculated as:
Time = 1.24 seconds.
The position of the ball when it is dropped from the cliff is calculated as:
Lastly, the distance below the top of the cliff is calculated as:
Distance = 7.53 meters.
Read more on distance here: brainly.com/question/10545161
Answer:
The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.
The production rate function M(t) is:
![m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min](https://tex.z-dn.net/?f=m%28t%29%3D%5BH%28t%29%5Ccdot10%2BH%28t-20%29%5Ccdot5-H%28t-80%29%5Ccdot14%2BH%28t-81%29%5Ccdot9%5Dkg%2Fmin)
(1)
The Laplace transform of this function is:
![\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%28s%29%3D%5B%5Cfrac%7B10%2B5e%5E%7B-20s%7D-14e%5E%7B-80s%7D%2B9e%5E%7B-81s%7D%7D%7Bs%7D%5Dkg%2Fmin)
(2)
Explanation:
The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.
For the Laplace transform we use the following rules:
![\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bf%28x%29%2Bg%28x%29%5D%3D%5Cmathcal%7BL%7D%5Bf%28x%29%5D%2B%5Cmathcal%7BL%7D%5Bg%28x%29%5D%3DF%28s%29%2BG%28s%29)
(3)
![\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5BaH%28x-b%29%5D%3D%5Cdisplaystyle%5Cfrac%7Bae%5E%7B-bs%7D%7D%7Bs%7D)
(4)
Answer:
Helps to remove Oxides formed during brazing.
Explanation:
Flux is needed to dissolve and remove oxides that may form during brazing. Prevent or inhibit the formation of oxide during the brazing process
Answer:
a. WATER CONTENT.
mass of water (MW)= 417g - 225g
Mw= 192g
M= (Mw / Ms)
M= [192/417]
M= 0.46
M= 46.0%
b. Void Ration (e)
To calculate the void ratio we must first calculate the volume of solids. Then we can find the volume of voids by subtracting the volume of solids from the total volume.
Ps= (Ms/Vs)=GsPw
Therefore
Vs= (Ms/GsPw)
Vs= (417/ 2.70*1000)
Vs= 0.154cm3
But V= Vv+Vs
Vv=V-Vs
e= (Vv/Vs)
Answer:
<u>Option 1</u>
Explanation:
As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :
- <u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>
