[ any unit of energy ] divided by [ any unit of time ]
is a perfectly good unit of power.
The most widely used one is [ joule / second ], called the [ <em>watt</em> ] .
Answer:
Position of object is;
s(t) = 4t³/3 + 3t + 1
Explanation:
We are told that the velocity has an expression;
v(t) = 3.00 m/s + ( 4.00 m/s³)t²
Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;
s(t) = ∫3 + 4t²
s(t) = 3t + 4t³/3 + c
Now, we were told that at x = 1.00 m, time t = 0.000 s
Thus, plugging the values in;
1 = 3(0) + 4(0³/3) + c
c = 1
Thus,the expression for the position of the object is;
s(t) = 4t³/3 + 3t + 1
Answer:
Fnet = 12 N
Explanation:
Force on a point charge due to another point charge = kq1q2 / d^2
Force on +32uC = due to + 20uC + due to -60uC
where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2
Net Force =
![= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%20_0%7D%20%5B%5Cfrac%7B32%20%5Ctimes%2010%5E-%5E6%20%5Ctimes60%5Ctimes10%5E-%5E6%7D%7B%2860%2F100%29%5E2%7D-%5Cfrac%7B32%20%5Ctimes%2010%5E-%5E6%20%5Ctimes20%5Ctimes10%5E-%5E6%7D%7B%20%2840%2F100%29%5E2%7D%20%20%5D)
![F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D9%20%5Ctimes10%5E9%5Ctimes%2010%5E-%5E1%5E2%5B%5Cfrac%7B32%5Ctimes60%5Ctimes10%5E4%7D%7B60%5Ctimes60%7D%20-%5Cfrac%7B32%5Ctimes20%5Ctimes10%5E4%7D%7B40%5Ctimes40%7D%20%5D)
![=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N](https://tex.z-dn.net/?f=%3D90%5B32%28%5Cfrac%7B80-60%7D%7B60%5Ctimes%2080%7D%20%29%5D%5C%5C%5C%5C%3D90%5Ctimes32%5Ctimes0.004167%5C%5C%5C%5C%3D12N)
Fnet = 12 N
We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.
As it leaves from rest, it means that the initial speed is zero
t₁=9 min=540 s
t₂=10 min=600 s
x₁=at₁²/2=8*540²/2=4*291600=1166400 m
x₂=at₂²/2=8*600²/2=4*360000=1440000 m
Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel
