Answer:
The correct option is;
Amplitude
Explanation:
When transmitting picture signals over the air by broadcasting stations, the signals are shifted into high frequency channels of Very High Frequency (VHF) or Ultra High Frequency (UHF) carrier currents and imposing the the television signal by changing the amplitude of the high frequency carrier current to match the transmitted television signal waveform shape
Answer: experiment data is the things you do in the experiment and the result is the answer
Answer:
20 [N], in the opposite direction of the first force.
Explanation:
We know that newton's second law stipulates that the sum of forces on a body must be equal to the product of mass by acceleration.
![SumF = m*a\\30 + F = 2*5\\F = 30 - (2*5)\\F = - 20 [N]](https://tex.z-dn.net/?f=SumF%20%3D%20m%2Aa%5C%5C30%20%2B%20F%20%3D%202%2A5%5C%5CF%20%3D%2030%20-%20%282%2A5%29%5C%5CF%20%3D%20-%2020%20%5BN%5D)
The negative sign means that the other force acting on the body must be in the opposite direction to the force of 30 [N]
Answer:
Option D
4200 W
Explanation:
Power, P is also given as the product of voltage and current, expressed as P=VI
Here, P is power, V is voltage in the xircuit and I is current theough voltage.
Taking 12 V for voltage across and 350A for current across circuit then power will be
P=350*12=4200 W
Therefore, option D is correct.
r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC