We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.
Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes
Therefore, the current flowing in the first device is
0.033 Amperes or 33 milliAmperes.
here since string is attached with a mass of 2 kg
so here tension force in the rope is given as
here we will have
now we will have speed of wave given as
here we will have
now we know that frequency is given as
F = 100 Hz
now wavelength is given as
so wavelength will be 0.16 m
Answer:
The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray. If a perpendicular were to be drawn on reflecting surface, it would be called normal. The figure below shows the reflection of an incident beam on a plane mirror.
Explanation:
Acceleration = (change in speed) / (time for the change)
change in speed = (speed at the end) - (speed at the beginning)
change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr
Acceleration = (-52 km/hr) / (6 sec)
Acceleration = (-26/3) km/(hr·sec)
Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²
(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)
= -26,000/(3 · 3600) m/s²
<em>Acceleration = -2.41 m/s²</em>
Answer:
I=V/R
so larger bsttery or small resistor can increase the current value.