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>
Answer:
Explanation:
Given that,
Initial speed of a car, u = 13 m/s
Final speed of a car, v = 25 m/s
Time, t = 5 s
We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :
So, the acceleration of the car is 
.
No, the rate of gravity remains constant
Answer:
Explanation:
Let the height of the ladder be L
Also:
- Let
- Let
When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.
So, we have:
--- Pythagoras Theorem
When the base is 9ft from the wall, this means that:
Substitute 9 for x and 10 for L in
Make 
the subject
Make y the subject
<em>Hence, the true distance at that point is approximately 4.36ft</em>
