Answer:
F = 2.25 N
Explanation:
Given that,
Charge 1, ![q_1=1\ \mu C = 10^{-6}\ C](https://tex.z-dn.net/?f=q_1%3D1%5C%20%5Cmu%20C%20%3D%2010%5E%7B-6%7D%5C%20C)
Charge 2, ![q_2=2.5\ \mu C=2.5\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_2%3D2.5%5C%20%5Cmu%20C%3D2.5%5Ctimes%2010%5E%7B-6%7D%5C%20C)
Distance between charges, d = 10 cm = 0.1 m
We need to find the electric force between charges. The formula that is used to find the electric force is given by
![F=\dfrac{kq_1q_2}{d^2}\\\\F=\dfrac{9\times 10^9\times 10^{-6}\times 2.5\times 10^{-6}}{(0.1)^2}\\\\=2.25\ N](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq_1q_2%7D%7Bd%5E2%7D%5C%5C%5C%5CF%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2010%5E%7B-6%7D%5Ctimes%202.5%5Ctimes%2010%5E%7B-6%7D%7D%7B%280.1%29%5E2%7D%5C%5C%5C%5C%3D2.25%5C%20N)
Hence, the force between charges is 2.25 N.
Answer:
A line crossing the x-axis in a velocity-time graph means that the moving object has changed its direction.
Explanation:
Velocity-Time graph:
A velocity-time graph is a two dimensional graph with velocity at its y-axis and time at its x-axis. At any point, value of y represents the velocity and value of x represents the time. The slope of the graph gives us the acceleration or deceleration of the moving object.
In a velocity-time graph:
- A straight line represents constant velocity.
- A diagonal line means that the velocity of a body is changing.
*Referring to the figure attached with the answer*
The velocity of the moving object increases at a constant rate for the first 10 minutes. Then the velocity is 60 m/min for the next 5 minutes. After that the velocity is decreasing. Till 30th minute when the velocity is at 0 m/min.
What happens here?
Velocity is a vector quantity. It has some direction. In a velocity-time graph, we are only concerned with two directions of velocity:
- Forward direction
- Backward direction
So, the object stops at 30th minute and starts moving in the reverse direction after that with an increasing velocity. <u>The point where the line cuts the x-axis is basically the point where the object starts moving in the reverse direction.</u>
The best ways to determine the safest driving speed are to 1) know the speed limit, and 2) maintain a safe following distance.<span> The safe following distance is to be enough distanced form the vehicle in front of you. But how much is enough?
A </span>driver<span> should ideally stay at least two seconds behind any vehicle that is directly in front of his or her vehicle.</span>
The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
![P=2\pi.R](https://tex.z-dn.net/?f=P%3D2%5Cpi.R)
Know recall that:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
We now need to know how much distance is traveled during a minute or 60 seconds:
![D=7200\times 2\pi\times R](https://tex.z-dn.net/?f=D%3D7200%5Ctimes%202%5Cpi%5Ctimes%20R)
Finally we divide this result with t=60 seconds:
![v=\frac{7200\times2\pi\times R}{60} \\ R=\frac{12}{2}=6 ](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B7200%5Ctimes2%5Cpi%5Ctimes%20R%7D%7B60%7D%0A%5C%5C%0AR%3D%5Cfrac%7B12%7D%7B2%7D%3D6%0A)
![v\approx 4523.89 \frac{units}{second}](https://tex.z-dn.net/?f=v%5Capprox%204523.89%20%5Cfrac%7Bunits%7D%7Bsecond%7D)
Where the distance units were named units as the length unit is not specified in this exercise.<span />
Answer:
Explanation:
Given
Potential Energy of a particle is given by ![U(x)=2x+\frac{8}{x}](https://tex.z-dn.net/?f=U%28x%29%3D2x%2B%5Cfrac%7B8%7D%7Bx%7D)
For minimum or maximum Potential Energy differentiate U(x) w.r.t x
![\frac{\mathrm{d} U}{\mathrm{d} x}=2-\frac{8}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20U%7D%7B%5Cmathrm%7Bd%7D%20x%7D%3D2-%5Cfrac%7B8%7D%7Bx%5E2%7D)
putting ![\frac{\mathrm{d} }{\mathrm{d} x}=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%3D0)
![2-\frac{8}{x^2}=0](https://tex.z-dn.net/?f=2-%5Cfrac%7B8%7D%7Bx%5E2%7D%3D0)
![x^2=\frac{8}{2}](https://tex.z-dn.net/?f=x%5E2%3D%5Cfrac%7B8%7D%7B2%7D)
![x^2=4](https://tex.z-dn.net/?f=x%5E2%3D4)
![x=\sqrt{4}=\pm 2](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B4%7D%3D%5Cpm%202)
To know minimum value check ![\frac{\mathrm{d^2} U}{\mathrm{d} x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%5E2%7D%20U%7D%7B%5Cmathrm%7Bd%7D%20x%5E2%7D)
at x=-2
![\frac{\mathrm{d^2} U}{\mathrm{d} x^2}=\frac{16}{x^3}=-\frac{16}{8}=-2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%5E2%7D%20U%7D%7B%5Cmathrm%7Bd%7D%20x%5E2%7D%3D%5Cfrac%7B16%7D%7Bx%5E3%7D%3D-%5Cfrac%7B16%7D%7B8%7D%3D-2)
so at x=-2 potential energy is minimum ![U=-8\ J](https://tex.z-dn.net/?f=U%3D-8%5C%20J)