Answer:
The value is
Explanation:
From the question we are told that
The maximized electric field strength is 
The diameter is 
Generally the cross -sectional area is mathematically represented as
=> 
=> 
=> 
Generally the maximum power is mathematically represented as
Here c is the speed of light with value 
is the permeability of free space with value 
=>
Answer:
I'm not sure
Explanation:
sorry I'm not smart that's why I have this app
Answer:
Coil 2 have 235 loops
Explanation:
Given
The number of loops in coil 1 is n
₁=
159
The emf induced in coil 1 is ε
₁
=
2.78
V
The emf induced in coil 2 is ε
₂
=
4.11
V
Let
n
₂ is the number of loops in coil 2.
Given, the emf in a single loop in two coils are same. That is,
ϕ
₁/n
₁=
ϕ
₂
n
₂⟹
2.78/159
=
4.11/
n
₂
n₂=
n₂=235
Therefore, the coil 2 has n
₂=
235 loops.
Answer:
Pebble A has 1/3 the acceleration as pebble B.
Explanation:
F = m×a
mass of a = 3 × mass of b (m_a = 3 × m_b)
Same starting force, F
m_a = mass of a
m_b = mass of b
a_a = acceleration of a
a_b = acceleration of b
F = m_a × a_a = m_b × a_b
3 × m_b × a_a = m_b × a_b
3 × a_a = a_b
OR
a_a = a_b / 3