I belive it could be 6.5 but I could be wrong
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:

The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) =
₀
V(r) =
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:

= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Answer:
it would take him 1 minute to run 304.5 meters and 1 second to run 5.075 meters
Explanation: