In a circuit having 2 lamps are connected in parallel to a battery
then the two lamps will be having the same potential as the battery
i.e

As per Ohm's law,
and 
In other words, each lamp's current is inversely related to its individual resistance. We only know the current in one of the bulbs in this specific instance. We would therefore need further information in order to calculate the current in the other light. Therefore, there isn't enough data to make a statement.
Under the assumption that all physical parameters, including temperature, remain constant, Ohm's law asserts that "the voltage across a conductor is directly proportional to the current flowing through it".
Learn more about Ohm's law here
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Answer:
Explanation:
Given that,
Frequency of radio signal is
f = 800kHz = 800,000 Hz.
Distance from transmitter
d = 8.5km = 8500m
Electric field amplitude
E = 0.9 V/m
The average energy density can be calculated using
U_E = ½•ϵo•E²
Where ϵo = 8.85 × 10^-12 F/m
Then,
U_E = ½ × 8.85 × 10^-12 × 0.9²
U_E = 3.58 × 10^-12 J/m²
The average electromagnetic energy density is 3.58 × 10^-12 J/m²
Answer:
beat frequency = 13.87 Hz
Explanation:
given data
lengths l = 2.00 m
linear mass density μ = 0.0065 kg/m
String A is under a tension T1 = 120.00 N
String B is under a tension T2 = 130.00 N
n = 10 mode
to find out
beat frequency
solution
we know here that length L is
L = n ×
........1
so λ =
and velocity is express as
V =
.................2
so
frequency for string A = f1 = 
f1 = 
f1 =
and
f2 =
so
beat frequency is = f2 - f1
put here value
beat frequency =
-
beat frequency = 13.87 Hz