Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:
![x=4t^3+3t^2-5t+2](https://tex.z-dn.net/?f=x%3D4t%5E3%2B3t%5E2-5t%2B2)
Where,
x is in meters and t is in sec
We know that,
Velocity,
![v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bdx%7D%7Bdt%7D%5C%5C%5C%5Cv%3D%5Cdfrac%7Bd%284t%5E3%2B3t%5E2-5t%2B2%29%7D%7Bdt%7D%5C%5C%5C%5Cv%3D12t%5E2%2B6t-5)
(a) i. t = 2 s
![v=12(2)^2+6(2)-5=55\ m/s](https://tex.z-dn.net/?f=v%3D12%282%29%5E2%2B6%282%29-5%3D55%5C%20m%2Fs)
At t = 4 s
![v=12(4)^2+6(4)-5=211\ m/s](https://tex.z-dn.net/?f=v%3D12%284%29%5E2%2B6%284%29-5%3D211%5C%20m%2Fs)
(b) Acceleration,
![a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7Bd%2812t%5E2%2B6t-5%29%7D%7Bdt%7D%5C%5C%5C%5Ca%3D24t%2B6)
Pu t = 3 s in the above equation
So,
![a=24(3)+6\\\\a=78\ m/s^2](https://tex.z-dn.net/?f=a%3D24%283%29%2B6%5C%5C%5C%5Ca%3D78%5C%20m%2Fs%5E2)
Hence, this is the required solution.
If the potential is given by v = xy - 3z-2, then the electric field has a y-component of X
When the charge is present in any form, a point in space has an electric field that is connected to it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.
Each location in space where a charge exists in any form can be considered to have an electric field attached to it. The electric force per unit charge is another name for an electric field. The electric field's equation is given as E = F / Q. Volts per meter (V/m) is the electric field's SI unit. Newton's per coulomb unit is the same as this one.
To learn more about electric field please visit-brainly.com/question/15800304
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Data:
![f_{2} = 42 Hz](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%2042%20Hz)
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%20%5Cfrac%7BnV%7D%7B2L%7D%20)
Wire 2 → 2º Harmonic → n = 2
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{2} = \frac{2V}{2L} ](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%20%5Cfrac%7B2V%7D%7B2L%7D%20%0A)
![2V = f_{2} *2L](https://tex.z-dn.net/?f=2V%20%3D%20%20f_%7B2%7D%20%2A2L)
![V = \frac{ f_{2}*2L }{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B%20f_%7B2%7D%2A2L%20%7D%7B2%7D%20)
![V = \frac{42*2L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B42%2A2L%7D%7B2%7D%20)
![V = \frac{84L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B84L%7D%7B2%7D%20)
![V = 42L](https://tex.z-dn.net/?f=V%20%3D%2042L)
Wire 1 → 1º Harmonic or Fundamental rope → n = 1
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{1} = \frac{1V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B1V%7D%7B2L%7D%20)
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%20%5Cfrac%7BV%7D%7B2L%7D%20)
If, We have:
V = 42L
Soon:
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7BV%7D%7B2L%7D%20)
![f_{1} = \frac{42L}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B42L%7D%7B2L%7D%20)
![\boxed{f_{1} = 21 Hz}](https://tex.z-dn.net/?f=%5Cboxed%7Bf_%7B1%7D%20%3D%2021%20Hz%7D)
Answer:
<span>The fundamental frequency of the string:
</span>
21 Hz
Answer:
E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.
Explanation:
#PLATOLIVESMATTER
Answer:
Waning Gibbous is correct