Answer:
A) 12.6 m/s2
B) -0.961 m
C) 0.961 m
D) 0.31 m/s
Explanation:
We can derive the acceleration function in term of t by taking 1st derivative of v(t)
![a(t) = \frac{dv(t)}{dt} = 6 - 6t](https://tex.z-dn.net/?f=a%28t%29%20%3D%20%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%20%3D%206%20-%206t)
The position function in term of t would be the integration of v(t)
![s(t) = \int {v(t)} \, dt=\int {(6t - 3t^2)} \, dt = 3t^2 - t^3 + s_0](https://tex.z-dn.net/?f=s%28t%29%20%3D%20%5Cint%20%7Bv%28t%29%7D%20%5C%2C%20dt%3D%5Cint%20%7B%286t%20-%203t%5E2%29%7D%20%5C%2C%20dt%20%3D%203t%5E2%20-%20t%5E3%20%2B%20s_0)
Since s = 0 when t = 0 we can conclude that
by plugging in t = 0 and s = 0
Part A: a(t = 3.1) = 6 - 6*3.1 = -12.6 m/s2. So the particle's deceleration is 12.6 m/s2 when t = 3.1
Part B:
![s(t = 3.1) = 3*3.1^2 - 3.1^3 = -0.961 m](https://tex.z-dn.net/?f=s%28t%20%3D%203.1%29%20%3D%203%2A3.1%5E2%20-%203.1%5E3%20%3D%20-0.961%20m%20)
Part C: As s(0) = 0 and s(3.1) = -0.961. The particle has traveled a distance of |-0.961 - 0| = 0.961m
Part D: So the particle has traveled a distance of 0.961m within time of 3.1s. That means the average speed is overall distance divided by overall time
= 0.961 / 3.1 = 0.31m/s
Answer:
20 N
Explanation:
The question above is related to "Newton's Third Law of Motion." It states that <em>every action is being opposed by a reaction of equal size but different direction</em>.
The action force in the situation above is 20 N while the reaction force here is the one coming from the volleyball. So if the action force is <em>20 N </em>then it's a common sense that the reaction force is also<em> 20 N.</em>
Thus, <u>20 N is the amount of force that the volleyball is exerting on Ana.</u>
C. element only one substance
Answer:
a. ![\alpha =2122.22\: rev/s^{2}](https://tex.z-dn.net/?f=%5Calpha%20%3D2122.22%5C%3A%20rev%2Fs%5E%7B2%7D)
b. ![\Delta \theta =9,550.02\: rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D9%2C550.02%5C%3A%20rev)
Explanation:
The computation is shown below:
data provided in the question
The initial angular velocity
= 0 rev/s,
f = Final angular velocity =
= 382000 rpm i.e =
= 6,366.67
And time = t = 3.0s
Based on the above information
a. For angular acceleration of drill
![\small \omega =\omega _{o}+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^{2}](https://tex.z-dn.net/?f=%5Csmall%20%5Comega%20%3D%5Comega%20_%7Bo%7D%2B%5Calpha%20t%20%5C%5C%5C%5C%5C%206%2C366.67%20%3D%200%20%2B%20%5Calpha%20%283.0%29%20%5C%5C%5C%5C%20%5Calpha%20%3D2122.22%5C%3A%20rev%2Fs%5E%7B2%7D)
b. For the number of revolutions
![\small \omega ^{2}-\omega _{o}^{2}=2\alpha \Delta \theta \\\\(6,366.67) ^{2}-(0)^{2}=2(2122.22) \Delta \theta \\\\ \Delta \theta =9,550.02\: rev](https://tex.z-dn.net/?f=%5Csmall%20%5Comega%20%5E%7B2%7D-%5Comega%20_%7Bo%7D%5E%7B2%7D%3D2%5Calpha%20%5CDelta%20%5Ctheta%20%5C%5C%5C%5C%286%2C366.67%29%20%5E%7B2%7D-%280%29%5E%7B2%7D%3D2%282122.22%29%20%5CDelta%20%5Ctheta%20%5C%5C%5C%5C%20%5CDelta%20%5Ctheta%20%3D9%2C550.02%5C%3A%20rev)
We simply applied the above formulas for determining each parts
Answer:
Acceleration, ![a=9.39\ m/s^2](https://tex.z-dn.net/?f=a%3D9.39%5C%20m%2Fs%5E2)
Explanation:
Given that,
Mass of the planet Krypton, ![m=8.8\times 10^{23}\ kg](https://tex.z-dn.net/?f=m%3D8.8%5Ctimes%2010%5E%7B23%7D%5C%20kg)
Radius of the planet Krypton, ![r=2.5\times 10^{6}\ m](https://tex.z-dn.net/?f=r%3D2.5%5Ctimes%2010%5E%7B6%7D%5C%20m)
Value of gravitational constant, ![G=6.6726\times 10^{-11}\ Nm^2/kg^2](https://tex.z-dn.net/?f=G%3D6.6726%5Ctimes%2010%5E%7B-11%7D%5C%20Nm%5E2%2Fkg%5E2)
To find,
The acceleration of an object in free fall near the surface of Krypton.
Solution,
The relation for the acceleration of the object is given by the below formula as :
![a=\dfrac{Gm}{r^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BGm%7D%7Br%5E2%7D)
![a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B6.6726%5Ctimes%2010%5E%7B-11%7D%5Ctimes%208.8%5Ctimes%2010%5E%7B23%7D%7D%7B%282.5%5Ctimes%2010%5E%7B6%7D%29%5E2%7D)
![a=9.39\ m/s^2](https://tex.z-dn.net/?f=a%3D9.39%5C%20m%2Fs%5E2)
So, the value of acceleration of an object in free fall near the surface of Krypton is ![9.39\ m/s^2](https://tex.z-dn.net/?f=9.39%5C%20m%2Fs%5E2)