Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:
![B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20r%7D%5C%5C%5C%5Cr%20%3D%20%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20B%7D)
where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,
![r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%284%5Cpi%5C%20x%5C%2010%5E%7B-7%7D%5C%20N%2FA%5E2%29%282%5C%20A%29%7D%7B2%5Cpi%288%5C%20x%5C%2010%5E%7B-6%7D%5C%20T%29%7D%5C%5C%5C%5C)
<u>r = 0.05 m = 5 cm</u>
Answer:
according to this question best answer is C
As per above given data
initial velocity = 19.3 km/s
final velocity = - 18.8 km/s
now in order to find the change in velocity
![\Delta v = v_f - v_i](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20v_f%20-%20v_i)
![\Delta v = -18.8 - 19.3](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20-18.8%20-%2019.3)
![\Delta v = -38.1 km/s](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20-38.1%20km%2Fs)
![\Delta v = -3.81 * 10^4 m/s](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20-3.81%20%2A%2010%5E4%20m%2Fs)
Part b)
Now we need to find acceleration
acceleration is given by formula
![a = \frac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
given that
![\Delta v =- 3.81 * 10^4 m/s](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D-%203.81%20%2A%2010%5E4%20m%2Fs)
![\Delta t = 2.07 years = 6.53 * 10^7 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%202.07%20years%20%3D%206.53%20%2A%2010%5E7%20s)
now the acceleration is given as
![a = \frac{-3.81 * 10^4}{6.53 * 10^7}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B-3.81%20%2A%2010%5E4%7D%7B6.53%20%2A%2010%5E7%7D)
![a = - 5.84 * 10^{-4}m/s^2](https://tex.z-dn.net/?f=a%20%3D%20-%205.84%20%2A%2010%5E%7B-4%7Dm%2Fs%5E2)
so above is the acceleration
Explanation:
There are five equations of motion:
v = at + v₀
Δx = v₀ t + ½ at²
Δx = ½ (v + v₀)t
v² = v₀² + 2aΔx
Δx = vt − ½ at²
Δx is the displacement
v₀ is the initial velocity
v is the final velocity
a is the acceleration
t is time