a. The disk starts at rest, so its angular displacement at time
is
![\theta=\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%5Calpha2t%5E2)
It rotates 44.5 rad in this time, so we have
![44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}](https://tex.z-dn.net/?f=44.5%5C%2C%5Cmathrm%7Brad%7D%3D%5Cdfrac%5Calpha2%286.00%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%5Calpha%3D2.47%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D)
b. Since acceleration is constant, the average angular velocity is
![\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Comega_f%2B%5Comega_i%7D2%3D%5Cdfrac%7B%5Comega_f%7D2)
where
is the angular velocity achieved after 6.00 s. The velocity of the disk at time
is
![\omega=\alpha t](https://tex.z-dn.net/?f=%5Comega%3D%5Calpha%20t)
so we have
![\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_f%3D%5Cleft%282.47%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%286.00%5C%2C%5Cmathrm%20s%29%3D14.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
making the average velocity
![\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B14.8%5Cfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%7D2%3D7.42%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
Another way to find the average velocity is to compute it directly via
![\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5CDelta%5Ctheta%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B44.5%5C%2C%5Crm%20rad%7D%7B6.00%5C%2C%5Crm%20s%7D%3D7.42%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
c. We already found this using the first method in part (b),
![\omega=14.8\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega%3D14.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
d. We already know
![\theta=\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%5Calpha2t%5E2)
so this is just a matter of plugging in
. We get
![\theta=179\,\mathrm{rad}](https://tex.z-dn.net/?f=%5Ctheta%3D179%5C%2C%5Cmathrm%7Brad%7D)
Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that
![\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D44.5%5C%2C%5Cmathrm%7Brad%7D%2B%5Cleft%2814.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%5Cright%29t%2B%5Cdfrac%5Calpha2t%5E2)
Then for
we would get the same
.
Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂
![t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7B72v_2%7D%7Bv_1-v_2%7D%5C%5C%5C%5Ct_1%20%3D%20%20%20%5Cfrac%7B72%2A5.5%7D%7B8.5-5.5%7D%5C%5C%5C%5Ct_1%20%3D%20132%20%5C%20s)
The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km
Would it be an open cluster
Answer:
The inter-molecular forces holding non-polar compounds together is low compare to that of polar compounds. Therefore, it will take less energy to break the bond for non-polar compounds and vice versa. That is why polar compounds have higher melting points than non-polar compounds.
Explanation: