Electromagnectic Waves Travel In A Vacuum
Answer: 1.55 x 10⁴ Nm²c^-1
Explanation: The electric flux, electric field intensity and area are related by the formulae below.
Φ= EAcosθ,
Where Φ= electric flux (Nm²c^-1)
E =electric field intensity (N/m²)
A = Area (m²)
θ= this is angle between the planar area and the magnetic flux
For our question E=3.80KN/c= 3800 N/c
A= 0.700 x 0.350= 0.245m²
θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).
Hence Φ= 3800 x 0.245 x cos(0)
= 3800 x 0.245 x 1 (value of cos 0° =1)
= 1.55 x 10⁴ Nm²c^-1
Thus the electric field is 1.55 x 10⁴ Nm²c^-1
Answer:
It is connected in series with the circuit
Explanation:
This is because to measure the current in the circuit, the current in the circuit has to flow through the ammeter. As such, the ammeter must be connected in series with the circuit so as to measure the current flowing through the circuit.
So, to measure the current flowing through a circuit with an ammeter, the ammeter must be connected in series with the circuit.
Answer:
27.1m/s
Explanation:
Given parameters:
Height of the building = 30m
Initial velocity = 12m/s
Unknown:
Final velocity = ?
Solution:
We apply one of the kinematics equation to solve this problem:
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 12² + (2 x 9.8 x 30)
v = 27.1m/s
