Answer:
Maximum emf = 5.32 V
Explanation:
Given that,
Number of turns, N = 10
Radius of loop, r = 3 cm = 0.03 m
It made 60 revolutions per second
Magnetic field, B = 0.5 T
We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

For maximum emf, 
So,

So, the maximum emf generated in the loop is 5.32 V.
Answer:
Hello there use something that looks like this
Explanation:
This is an accurate representation of something you are working on!
As you can see the wire and the core are represented on the left and is showing how it can be represented on your right hand and how they are similar!
Answer:
7500 m/s
Explanation:
Centripetal acceleration = gravity
v² / r = GM / r²
v = √(GM / r)
Given:
G = 6.67×10⁻¹¹ m³/kg/s²
M = 5.98×10²⁴ kg
r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m
v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))
v = 7500
The orbital velocity is 7500 m/s.
Answer:
v = 302.923 m/s
Explanation:
We can answer this question using conservation of energy. Since there is no energy loss (e.g. no viscous drag) the energy when they are far apart and the energy when they barely touch must be the same.
The initial energy must be equal to the sum of their kinetic energies, since they are far apart to feel their electrical interaction.
Ei = (1/2)mv1^2 + (1/2)m*v2^2
Let us consider that they move with the same speed:
Ei = mv^2
If we consider the case when they barely touch, there won't be any kinetic energy, just pure electromagnetic energy:
Ef = k q1q2/(r1+r2) = k q1q2/(2r1)
Since Ei = Ef
v^2 = (k/m) q1q2/(2r1)
where
k = 8.98755 x10^9 Nm^2/C^2
m = 9.05 x10^-14 kg
q1 = −2.10 pC
q2 = −3.30 pC
r1 = 3.75×10^−6 m
v^2 = 91762.4 m^2/s^2
v = 302.923 m/s
Answer:
a)
, b) 
Explanation:
a) The equation for vertical velocity is obtained by deriving the function with respect to time:

The velocities at given instants are, respectivelly:

