Answer:
0.00000609180381907 T
Explanation:
= Vacuum permeability = ![4\pi \times 10^{-7}\ H/m](https://tex.z-dn.net/?f=4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5C%20H%2Fm)
I = 3.95 A
r = 6.05 cm
Cylinder area
![a=\pi(0.0705^2-0.0465^2)](https://tex.z-dn.net/?f=a%3D%5Cpi%280.0705%5E2-0.0465%5E2%29)
Area within r = 6.05 cm
![A=\pi(0.0605^2-0.0465^2)](https://tex.z-dn.net/?f=A%3D%5Cpi%280.0605%5E2-0.0465%5E2%29)
Current in the enclosure is given by
![I_1=\dfrac{A}{a}I\\\Rightarrow I_1=\dfrac{\pi(0.0605^2-0.0465^2)}{\pi(0.0705^2-0.0465^2)}\times 3.95\\\Rightarrow I_1=2.10722934473\ A](https://tex.z-dn.net/?f=I_1%3D%5Cdfrac%7BA%7D%7Ba%7DI%5C%5C%5CRightarrow%20I_1%3D%5Cdfrac%7B%5Cpi%280.0605%5E2-0.0465%5E2%29%7D%7B%5Cpi%280.0705%5E2-0.0465%5E2%29%7D%5Ctimes%203.95%5C%5C%5CRightarrow%20I_1%3D2.10722934473%5C%20A)
Net enclosed current
![I_n=I-I_1\\\Rightarrow I_n=3.95-2.10722934473\\\Rightarrow I_n=1.84277065527\ A](https://tex.z-dn.net/?f=I_n%3DI-I_1%5C%5C%5CRightarrow%20I_n%3D3.95-2.10722934473%5C%5C%5CRightarrow%20I_n%3D1.84277065527%5C%20A)
Magnetic field is given by
![B=\dfrac{\mu_0I_n}{2\pi r}\\\Rightarrow B=\dfrac{4\pi \times 10^{-7}\times 1.84277065527}{2\pi\times 0.0605}\\\Rightarrow B=0.00000609180381907\ T](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_0I_n%7D%7B2%5Cpi%20r%7D%5C%5C%5CRightarrow%20B%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%201.84277065527%7D%7B2%5Cpi%5Ctimes%200.0605%7D%5C%5C%5CRightarrow%20B%3D0.00000609180381907%5C%20T)
The magnetic field strength is 0.00000609180381907 T
-- IF the meter stick is oriented perpendicular to the direction from
it to you, then it appears 1 meter long to you, but very very skinny.
The Lorentz contraction only applies in the direction of motion.
-- If the length of the meter stick is pointing toward you, then . . .
L = L₀ √ (1 - v²/c²)
= √ (1 - 0.36)
= √ (0.64) = 0.8 meter .
To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first
![F_1 = F \frac{{343}}{(343-V)}](https://tex.z-dn.net/?f=F_1%20%3D%20F%20%5Cfrac%7B%7B343%7D%7D%7B%28343-V%29%7D)
Where F is the actual frequency and v is the velocity of the ambulance
Now the source is moving away from the observer.
![F_2 = F\frac{343}{(343+V)}](https://tex.z-dn.net/?f=F_2%20%3D%20F%5Cfrac%7B343%7D%7B%28343%2BV%29%7D)
We are also so told the perceived frequency decreases by 11.9%
![F_2 = F_1 - 9.27\% \text{ of } F_1](https://tex.z-dn.net/?f=F_2%20%3D%20F_1%20-%209.27%5C%25%20%5Ctext%7B%20of%20%7D%20F_1)
![F_2 = F_1-0.0927F_1](https://tex.z-dn.net/?f=F_2%20%3D%20F_1-0.0927F_1)
![F_2 = 0.9073F_1](https://tex.z-dn.net/?f=F_2%20%3D%200.9073F_1)
Equating,
![F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})](https://tex.z-dn.net/?f=F%5Cfrac%7B343%7D%7B%28343%2BV%29%7D%3D%200.9073%28F%5Cfrac%7B343%7D%7B%28343-V%29%7D%29)
![\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28343%2BV%29%7D%3D%200.9073%5Cfrac%7B1%7D%7B%28343-V%29%7D)
![0.9073(343+V) = 343-V](https://tex.z-dn.net/?f=0.9073%28343%2BV%29%20%3D%20343-V)
![(0.9073)(343)+(0.9073)V = 343-V](https://tex.z-dn.net/?f=%280.9073%29%28343%29%2B%280.9073%29V%20%3D%20343-V)
![V+0.9073V = 343-(0.9073)(343)](https://tex.z-dn.net/?f=V%2B0.9073V%20%3D%20343-%280.9073%29%28343%29)
Solving for V,
![V = 16.67 m/s](https://tex.z-dn.net/?f=V%20%3D%2016.67%20m%2Fs)
The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:
Vf = Vi + at
where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
Substituting the values gives:
30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
Answer:
L = 2.83 J.s
Explanation:
The formula for the angular momentum of the stone is given as follows:
L = mvr
where,
L = angular momentum of the stone = ?
m = mass of the stone = 0.1 kg
v = linear velocity of the stone = rω
r = radius of circular path = 1.5 m
ω = angular speed of the stone = (2 rev/s)(2π rad/1 rev) = 4π rad/s
Therefore,
L = mvr = m(rω)r
L = mr²ω
using values, we get:
L = (0.1 kg)(1.5 m)²(4π rad/s)
<u>L = 2.83 J.s</u>