Answer;
B. Fluorescent lamps operate at a higher temperature than incandescent lamps.
Explanation;
-A fluorescent lamp, is a type of electric light (lamp) that uses ultraviolet emitted by mercury vapor to excite a phosphor, which emits visible light.
-A fluorescent lamp produces less heat, thus, it is much more efficient. A fluorescent bulb can produce between 50 and 100 lumens per watt. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs.
-Fluorescent lamps operate best around room temperature. At much lower or higher temperatures, efficacy decreases.
Step 1: Identify the variables. ...Step 2: Determine the variable range. ...Step 3: Determine the scale of the graph. ...Step 4: Number and label each axis and title the graph.Step 5: Determine the data points and plot on the graph. ...Step 6: Draw the graph.
Answer:
Explanation:
Given
Frequency of SHM is 
Amplitude of SHM is 
Cup begins to slip when it overcomes the friction force
Friction force 
Applied force 
and maximum acceleration during SHM is
Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Answer:
a. Zin = 41.25 - j 16.35 Ω
b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c. Pin = 216 w
d. PL = Pin = 216 w
e. Pg = 478.4 w , Pzg = 262.4 w
Explanation:
a.
Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]
βl = 2π / λ * 0.15 λ = 54 °
Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]
Zin = 41.25 - j 16.35 Ω
b.
I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶
V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)
V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c.
Pin = ¹/₂ * Re * [V₁ * I₁]
Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)
Pin = 216 w
d.
The power PL and Pin are the same as the line is lossless input to the line ends up in the load so
PL = Pin
PL = 216 w
e.
Pg Generator
Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)
Pg = 478.4 w
Pzg dissipated
Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50
Pzg = 262.4 w
