<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer:
1.90×10²⁰ Electrons
Explanation:
From the question,
Q = It.................... Equation 1
Where Q = charge flowing through the wire, I = current, t = time
Given: I = 4.35 A, t = 7.00 s
Substitute these values into equation 1
Q = 4.35(7.00)
Q = 30.45 C.
But,
1 electron contains 1.6×10⁻¹⁹ C
therefore,
30.45 C = 30.45/1.6×10⁻¹⁹ electrons
= 1.90×10²⁰ Electrons
Answer: c
Explanation: because if it increases then it gets hotter
if it decreases then its getting colder.
