The I3 will be 158 A.
<h3>How to find the current through the circuit?</h3>
- The foundation of circuit analysis is Kirchhoff's circuit laws.
- We have the fundamental instrument to begin studying circuits with the use of these principles and the equation for each individual component (resistor, capacitor, and inductor).
- These rules aid in calculating the current flow in various network streams as well as the electrical resistance of a complicated network, or impedance in the case of AC.
To calculate I3 firstly, V4 has to be calculated,
For I3,
Hence, the current through I3 will be 158 A.
To learn more about Kirchoff's laws refer to:
brainly.com/question/86531
#SPJ 10
Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.
<h3><u>
Explanation:</u></h3>
The range in which the light exists is described as the electromagnetic spectrum. The light waves, radio waves, gamma rays,etc that exist in the world is not visible to human eyes. A kind of wave that modifies magnetic and electric fields is light. Spectroscopy makes use of all the frequencies and the wavelengths of the electromagnetic radiation.
The part of the electromagnetic spectrum that can be seen by the human eyes is the visible spectrum. The light waves with the wavelengths of 380 to 740 nm can be sen by the human eyes. Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.
Answer:
a.Internal temperature change is 90.49°F
b.Internal energy change is -2.43Btu/lbm
Explanation:
The energy balance for this steady-flow system can be expressed as:
Therefore,
#Since:
From tableA-1
#Replacing the values in the equation:
We have that
0.0024
Milimetres are before centimetres and centimetres are before metres
Given that:
spring compressed ( x) = 0.2 m
Work (W) = 24 J
determine , spring constant (x)=?
We know hat ,
Work (W) = (1/2). k.x²
24 = 1/2 ( k × 0.2²)
48 = k × 0.2²
<em> k = 1200 N/m</em>