<span>Two characteristics of regular, periodic waveforms are :
</span><span>1) Amplitude - It is the </span><span>the length and width of waves, such as sound
waves, as they move or vibrate. An example would be how
much a radio wave moving back and forth.
</span><span>
2) Frequency - It is the number of waves cycles per unit of time, passing a
point per unit time. It is usually measured in Hertz.</span>
Answer:
1. 12 V
2a. R₁ = 4 Ω
2b. V₁ = 4 V
3a. A = 1.5 A
3b. R₂ = 4 Ω
4. Diagram is not complete
Explanation:
1. Determination of V
Current (I) = 2 A
Resistor (R) = 6 Ω
Voltage (V) =?
V = IR
V = 2 × 6
V = 12 V
2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:
Voltage (V) = 12 V
Current (I) = 1 A
Equivalent resistance (R) =?
V = IR
12 = 1 × R
R = 12 Ω
a. Determination of R₁
Equivalent resistance (R) = 12 Ω
Resistor 2 (R₂) = 8 Ω
Resistor 1 (R₁) =?
R = R₁ + R₂ (series arrangement)
12 = R₁ + 8
Collect like terms
12 – 8 =
4 = R₁
R₁ = 4 Ω
b. Determination of V₁
Current (I) = 1 A
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 1 × 4
V₁ = 4 V
3a. Determination of the current.
Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) = 6 V
Current (I) =?
V₁ = IR₁
6 = 4 × I
Divide both side by 4
I = 6 / 4
I = 1.5 A
Thus, the ammeter (A) reading is 1.5 A
b. Determination of R₂
We'll begin by calculating the voltage cross R₂. This can be obtained as follow:
Total voltage (V) = 12 V
Voltage 1 (V₁) = 6 V
Voltage 2 (V₂) =?
V = V₁ + V₂ (series arrangement)
12 = 6 + V₂
Collect like terms
12 – 6 = V₂
6 = V₂
V₂ = 6 V
Finally, we shall determine R₂. This can be obtained as follow:
Voltage 2 (V₂) = 6 V
Current (I) = 1.5 A
Resistor 2 (R₂) =?
V₂ = IR₂
6 = 1.5 × R₂
Divide both side by 1.5
R₂ = 6 / 1.5
R₂ = 4 Ω
4. The diagram is not complete
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
To solve this problem it is necessary to apply the concepts related to simple harmonic movement.
The maximum speed from the simple harmonic motion is given as
![V = A\sqrt{\frac{K}{m}}](https://tex.z-dn.net/?f=V%20%3D%20A%5Csqrt%7B%5Cfrac%7BK%7D%7Bm%7D%7D)
Where,
K = Spring constant
m = mass
At this case m is a constant then
![V \propto \sqrt{K}](https://tex.z-dn.net/?f=V%20%5Cpropto%20%5Csqrt%7BK%7D)
then the ratio is given by
![\frac{v_2}{v_1}=\sqrt{\frac{K_1}{K_2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_2%7D%7Bv_1%7D%3D%5Csqrt%7B%5Cfrac%7BK_1%7D%7BK_2%7D%7D)
According the statement,
![v_2 = \sqrt{\frac{K_1}{K_2}}v_1](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5Cfrac%7BK_1%7D%7BK_2%7D%7Dv_1)
![v_2 = 2v_1](https://tex.z-dn.net/?f=v_2%20%3D%202v_1)
Therefore the maximum speed becomes double: 2) It would increase by a factor of 2.