Answer:
Speed or velocity is plotted on the Y-axis. A straight horizontal line on a speed-time graph means that speed is constant. It is not changing over time. A straight line does not mean that the object is not moving
Acceleration = (change in speed) / (time for the change).
Acceleration = (14 m/s - 0 m/s) / (3.5 s) = <u>4 m/s²</u>
Answer:
a)![v=\dfrac{2.KE}{qBR}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2.KE%7D%7BqBR%7D)
b)![m=\dfrac{(qBR)^2}{2.KE}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B%28qBR%29%5E2%7D%7B2.KE%7D)
Explanation:
Given that
Charge = q
Magnetic filed = B
Radius = R
We know that kinetic energy KE
----------1
m v² = 2 .KE
The magnetic force F = q v B
Radial force
![Fr=\dfrac{1}{R}mv^2](https://tex.z-dn.net/?f=Fr%3D%5Cdfrac%7B1%7D%7BR%7Dmv%5E2)
For uniform force these two forces should be equal
![q v B=\dfrac{1}{r}mv^2](https://tex.z-dn.net/?f=q%20v%20B%3D%5Cdfrac%7B1%7D%7Br%7Dmv%5E2)
q v B R =m v²
q v B R = 2 .KE
![v=\dfrac{2.KE}{qBR}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2.KE%7D%7BqBR%7D)
Now put the velocity v in the equation
![m=\dfrac{2.KE}{\left(\dfrac{2.KE}{qBR}\right)^2}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B2.KE%7D%7B%5Cleft%28%5Cdfrac%7B2.KE%7D%7BqBR%7D%5Cright%29%5E2%7D)
![m=\dfrac{(qBR)^2}{2.KE}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B%28qBR%29%5E2%7D%7B2.KE%7D)
Answer:
Change in momentum is ![2.667\times 10^{4} kg.m/s](https://tex.z-dn.net/?f=2.667%5Ctimes%2010%5E%7B4%7D%20kg.m%2Fs)
Solution:
The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:
![\vec{p} = m\vec{v}](https://tex.z-dn.net/?f=%5Cvec%7Bp%7D%20%3D%20m%5Cvec%7Bv%7D)
Now, as per the question:
Mass of the car, M = 1500 kg
The velocity in the east direction, ![v\hat{i} = 40\hat{i} km/h](https://tex.z-dn.net/?f=v%5Chat%7Bi%7D%20%3D%2040%5Chat%7Bi%7D%20km%2Fh)
The velocity in the north direction, ![v\hat{j} = 50\hat{j} km/h](https://tex.z-dn.net/?f=v%5Chat%7Bj%7D%20%3D%2050%5Chat%7Bj%7D%20km%2Fh)
Now, the momentum of the car in the east direction:
![p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h](https://tex.z-dn.net/?f=p%5Chat%7Bi%7D%20%3D%20mv%5Chat%7Bi%7D%20%3D%201500%5Ctimes%2040%5Chat%7Bi%7D%20%3D%2060000%5Chat%7Bi%7D%20kg.km%2Fh)
Now, the momentum of the car in the north direction:
![p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h](https://tex.z-dn.net/?f=p%5Chat%7Bj%7D%20%3D%20mv%5Chat%7Bj%7D%20%3D%201500%5Ctimes%2050%5Chat%7Bj%7D%20%3D%2075000%5Chat%7Bj%7D%20kg.km%2Fh)
Change in momentum is given by:
![\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20p%5Chat%7Bi%7D%20-%20p%5Chat%7Bj%7D%20%3D%2060000%5Chat%7Bi%7D%20-%2075000%5Chat%7Bj%7D)
Now,
![|\Delta p| = |60000\hat{i} - 75000\hat{j}|](https://tex.z-dn.net/?f=%7C%5CDelta%20p%7C%20%3D%20%7C60000%5Chat%7Bi%7D%20-%2075000%5Chat%7Bj%7D%7C)
![|\Delta p| = \sqrt{60000^{2} + 75000^{2}}](https://tex.z-dn.net/?f=%7C%5CDelta%20p%7C%20%3D%20%5Csqrt%7B60000%5E%7B2%7D%20%2B%2075000%5E%7B2%7D%7D)
![|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s](https://tex.z-dn.net/?f=%7C%5CDelta%20p%7C%20%3D%2096046%20kg.km%2Fhr%20%3D%20%5Cfrac%7B96046%7D%7B3.6%7D%20%3D%202.667%5Ctimes%2010%5E%7B4%7D%20kg.m%2Fs)
(Since,
)
Because it is the force exerted on an object under normal conditions.