Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
Answer:
Explanation:
The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.
The strength of electric field is defined as the force experienced by the unit positive test charge.
E = F / q
Electric field strength is a vector quantity and it is measured in newton per coulomb.
Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.
The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.
Explanation:
Given that,
Capacitor 
Resistor 
Peak voltage = 5.10 V
(A). We need to calculate the crossover frequency
Using formula of frequency

Where, R = resistor
C = capacitor
Put the value into the formula


(B). We need to calculate the
when 
Using formula of 

Put the value into the formula


(C). We need to calculate the
when 
Using formula of 


(D). We need to calculate the
when 
Using formula of 


Hence, This is the required solution.
<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
</span>
Answer:
567.321nm
Explanation:
See attached handwritten document for more details