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:
C the particle must be somewhere.
Explanation:
This is because normalization of wave function means the maximum probability of finding a particle in a region is 1. And a Wave function describes the probability of finding a particle in region. Also Since it is a probability distribution, its integral over all space must be 1, explaining that the probability that the particle is somewhere and thus it must integrate to 1, meaning it must be it must be normalizable
Boiling point and acc motivation
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer:
While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>
Explanation:
hope this helps!