A must be at least 4 full paragraphs probably will need more
Answer:
I₁ = 1.6 A (through 7 Ohm Resistor)
I₂ = 1.3 A (through 8 Ohm Resistor)
I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)
Explanation:
Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:
R₁ = 7 Ω
R₂ = 4 Ω
R₃ = 8 Ω
V₁ = 12 V
V₂ = 9 V
Now, we apply KVL to 1st Loop:
V₁ = I₁R₁ + (I₁ - I₂)R₂
12 = 7I₁ + (I₁ - I₂)(4)
12 = 7I₁ + 4I₁ - 4I₂
I₁ = (12 + 4 I₂)/11 ------------ equation (1)
Now, we apply KVL to 2nd Loop:
V₂ = (I₂ - I₁)R₂ + I₂R₃
9 = (I₂ - I₁)(4) + 8I₂
9 = 4I₂ - 4I₁ + 8I₂
9 = 12I₂ - 4I₁ -------------- equation (2)
using equation (1)
9 = 12I₂ - 4[(12 + 4 I₂)/11]
99 = 132 I₂ - 48 - 16 I₂
147 = 116 I₂
I₂ = 147/116
I₂ = 1.3 A
use this value in equation 2:
9 = 12(1.3 A) - 4I₁
4I₁ = 15.6 - 9
I₁ = 6.6 A/4
I₁ = 1.6 A
Hence, the currents through all resistors are:
<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>
<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>
<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>
<h2><u>
Answer:</u></h2>
<u></u>
<u>Force = kgm/s² ⟶ Newton</u>
<h2><u>
Explanation:</u></h2>
<u></u>
According to the formula we've learnt,
- Force = mass × acceleration.
To find force in terms of base units, we must first identify the base SI units of mass & acceleration.
- Base SI unit of mass = kg (kilogram)
- Base SI unit of acceleration = m/s² (meter per second squared)
So, the base unit of force is,
- Force = mass × acceleration.
- <u>Force = kgm/s² ⟶ Newton</u>
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Hope it helps!
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Answer:the 90% is spent performing other functions
Explanation:
