Answer:
(1) 2 ohms
(2) 12 Volts Across each resistor
(3) I₁ = 3 A, I₂ = 2 A, I₃ = 1 A
Explanation:
From the question,
(1) Equilvalent Resistance (Rt) for parallel connection is
1/Rt =(1/R₁)+ (1/R₂) + (1/R₃)
Where R₁ = 4 ohms, R₂ = 6 ohms, R₃ = 12 ohms
1/Rt = 1/4 +1/6 +1/12
1/Rt = (3+2+1)/12
1/Rt = 6/12
1/Rt = 1/2
Rt = 2 ohms.
(2) Since the resistors are connected in parallel, They will have the same potential difference across them,
Hence the P.d across each resistor = 12 Volts.
(3) For R₁,
I₁ = V/R₁ = 12/4
I₁ = 3 A.
For R₂,
I₂ = V/R₂
I₂ = 12/6
I₂ = 2 A
For R₃,
I₃ = V/R₃
I₃ = 12/12
I₃ = 1 A
Answer:
The frequency of oscillation of the simple pendulum is 0.49 Hz.
Explanation:
Given that,
Mass of the simple pendulum, m = 0.35 kg
Length of the string to which it is attached, l = 1 m
We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :
So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.
Answer:
(a)
(b)
Explanation:
It is given that,
Force acting on the particle, F = 12 N
Displacement of the particle,
Magnitude of displacement,
(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :
is the angle between force and the displacement
According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.
So,
(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :
Hence, this is the required solution.
B. How much work can be done in a given time. That’s why it’s measured occasionally I. “Horsepower.” It’s your ability to work fast and far.
Answer:
They use noise control, creating a wave that negates outside or ambient sound and replaces it with the desired sound that listeners request.
Explanation:
I hope this helped