Answer:
This is how I figured it out:
- 215.5 rounded to one significant figure is 200
- 101.02555 rounded to one significant figure is 100.
- 200 + 100 = 300.
Hope this helps!
Explanation:
Explanation:
It is given that, the position of a particle as as function of time t is given by :

Let v is the velocity of the particle. Velocity of an object is given by :

![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)

So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :

![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)

At t = 0, 

Hence, this is the required solution.
b) between poles M1 and M2
Explanation:
From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.
The law of attraction between two magnetic poles states that:
<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s
Mathematically:
FM = K 
here r is the distance between the poles
FM is the magnetic force between the poles
M1 is the strength of the first magnetic pole
M2 is the strength of the second pole
K is the magnetic field constant
learn more:
magnetic pole brainly.com/question/2191993
#learnwithBrainly
Okay, first off, the formula for Kinetic Energy is:
<em>KE = 1/2(m)(v)^2</em>
<em>m = mass</em>
<em>v = velcoity (m/s)</em>
Using this formula, we can then calculate the kinetic energy in each scenario:
1) KE = 1/2(100)(5)^2 = 1,250 J
2) KE = 1/2(1000)(5)^2 = 12,500 J
3) KE = 1/2(10)(5)^2 = 125 J
4) KE = 1/2(100)(5)^2 = 1,250 J
Answer:
After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+
)
Explanation:
I believe it is a part C question.
The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........
Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.