Answer:
The output power of the circuit is 3 Watts.
Given:
a loss in decibels = 3 dB
Input power = 6 Watts
To find:
What is the output power?
Formula used:
Output power = Input power × loss in ratio
Solution:
3 dB loss corresponds to a ratio of 0.5
Output power can be calculated as follows:
Output power = Input power × loss in ratio
Output power = 6 × 0.5
Output power = 3 Watts
Therefore, the output power of the circuit is 3 Watts.
Answer:
a) Blood mass is 
b) The count of blood cells is 
Explanation:
From the problem statement, we learn that
The blood volume is 
The density of blood is 
% of blood which consists of cells is = 45.0%
the % of blood that is plasma is = 55.0%
density of blood cells is 
% of cells that are white is = 1%
% of cells that are red is = 99%
The red blood cell diameter is = 
The red blood cell radius is 
The mass is generally represented mathematically as
Substituting values
Cell mass is 
= 45% of m
= 0.45 * 5,7876
= 2.60442 kg
The volume of cells is
=
The white blood cells volume is 
= 1% of the cells volume
The volume of a single cell is 
The red blood cells volume is 
The total red blood cell count is 
The total white blood cell count is 
The overall number of blood cells is 
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
Δd = 23 cm. When the eta string of the guitar has nodes at both ends, the resulting waves create a standing wave, which can be expressed with the following formulas: Fundamental: L = ½ λ, 1st harmonic: L = 2 ( λ / 2), 2nd harmonic: L = 3 ( λ / 2), Harmonic n: L = n λ / 2, where n is an integer. The rope's speed can be calculated using the formula v = λ f. This speed remains constant based on the tension and linear density of the rope. Now, let's determine the speed with the provided data: v = 0.69 × 196, yielding v = 135.24 m/s. Next, we will find the wavelengths for the two frequencies: λ₁ = v / f₁, which gives λ₁ = 135.24 / 233.08, equaling λ₁ = 0.58022 m; λ₂ = v / f₂ results in λ₂ = 135.24 / 246.94, consequently λ₂ = 0.54766 m. We'll substitute into the resonance equation Lₙ = n λ/2. At the third fret, m = 3, therefore L₃ = 3 × 0.58022 / 2, resulting in L₃ = 0.87033 m. For the fourth fret, m = 4, which gives L₄ = 4 × 0.54766 / 2, equating to L₄ = 1.09532 m. The distance between the two frets is Δd = L₄ – L₃, so Δd = 1.09532 - 0.87033, leading to Δd = 0.22499 m or 22.5 cm, rounded to 23 cm.
25.82 m/s
Explanation:
Given:
Force applied by the baseball player; F = 100 N
Distance the ball travels; d = 0.5 m
Mass of the ball; m = 0.15 kg
To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved.
It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows:
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
Solving gives:
v² = (2 × 100 × 0.5) / 0.15
v² = 666.67
v = √666.67
v = 25.82 m/s.
