Answer:
Mass of ion will be
Explanation:
We have given ion is triply charged that is ![q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C](https://tex.z-dn.net/?f=q%3D3%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%3D4.8%5Ctimes%2010%5E%7B-19%7DC)
Radius r = 36 cm = 0.36 m
Velocity of the electron ![v=7\times 10^6m/sec](https://tex.z-dn.net/?f=v%3D7%5Ctimes%2010%5E6m%2Fsec)
Magnetic field B = 0.55 T
We know that radius of the path is given by ![r=\frac{mv}{qB}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bmv%7D%7BqB%7D)
![m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7BrqB%7D%7Bv%7D%3D%5Cfrac%7B0.36%5Ctimes%204.8%5Ctimes%2010%5E%7B-19%7D%5Ctimes%207%5Ctimes%2010%5E6%7D%7B0.55%7D%3D22%5Ctimes%2010%5E%7B-13%7Dkg)
Answer:
![v_0 = 3.53~{\rm m/s}](https://tex.z-dn.net/?f=v_0%20%3D%203.53~%7B%5Crm%20m%2Fs%7D)
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
![x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}](https://tex.z-dn.net/?f=x%20-%20x_0%20%3D%20v_%7Bx_0%7D%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_x%20t%5E2%5C%5C63%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20v_0t%20%2B%200%5C%5Ct%20%3D%20%5Cfrac%7B63%5Ctimes%2010%5E%7B-3%7D%7D%7Bv_0%7D)
For the y-direction gives
![v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t](https://tex.z-dn.net/?f=v_y%20%3D%20v_%7By_0%7D%20%2B%20a_y%20t%5C%5Cv_y%20%3D%200%20-9.8t%5C%5Cv_y%20%3D%20-9.8t)
Combining both equation yields the y_component of the final velocity
![v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}](https://tex.z-dn.net/?f=v_y%20%3D%20-9.8%28%5Cfrac%7B63%5Ctimes%2010%5E%7B-3%7D%7D%7Bv_0%7D%29%20%3D%20-%5Cfrac%7B0.61%7D%7Bv_0%7D)
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
![\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_x%7D%5C%5C%5Ctan%28177.2%5E%5Ccirc%29%20%3D%20-0.0489%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_0%7D%20%3D%20%5Cfrac%7B-0.61%2Fv_0%7D%7Bv_0%7D%20%3D%20-%5Cfrac%7B0.61%7D%7Bv_0%5E2%7D%5C%5Cv_0%20%3D%203.53~%7B%5Crm%20m%2Fs%7D)
Answer:21.18 m
Explanation:
Given
initial speed u=10 m/s
height of building h=22 m
time taken to complete 22 m
![h=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
initial vertical velocity =0
![22=\frac{1}{2}gt^2](https://tex.z-dn.net/?f=22%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![t=\sqrt{\frac{22\times 2}{g}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B22%5Ctimes%202%7D%7Bg%7D%7D)
![t=2.11 s](https://tex.z-dn.net/?f=t%3D2.11%20s)
Horizontal Distance moved
![R=u_x\times t](https://tex.z-dn.net/?f=R%3Du_x%5Ctimes%20t)
![R=10\times 2.11](https://tex.z-dn.net/?f=R%3D10%5Ctimes%202.11)
![R=21.18 m](https://tex.z-dn.net/?f=R%3D21.18%20m)
Answer:
The momentum of an object is defined as the mass of the object times the velocity of the object, as P = m*v.
So the equipment needed would be:
Something to measure the mass of the object, like a balance.
Something to measure the speed of the object, like a doppler radar, or a simpler thing may be a cronometer, with that you can measure the amount of time that the object needs to travel a given distance, and with that you can obtain the speed of the object.
Now you can notice that speed is different than velocity, this is true, velocity is a vector, so this has a direction, then you need something to fix the direction in which the object moves, in this way you can determine the velocity.