Answer:
- The total distance traveled is 28 inches.
- The displacement is 2 inches to the east.
Explanation:
Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector 
pointing in the west direction, the ant start at position
Then, moves to
so, the distance traveled here is
after this, the ant travels to
so, the distance traveled here is
The total distance traveled will be:
The displacement is the final position vector minus the initial position vector:
This is 2 inches to the east.
Answer:
5702.88 J or 5.7mJ
Explanation:
Given that :
C 1 = 6.0-μF
C 2 = 4.0-μF
V 1 = 50V
V 2 = 34V
Note that : Q = CV
Q 1 = C1 * V1
Q 1 = 50×6 = 300μC
Q 2 = 34×4 = 136μC
Parallel connection = C 1 + C 2
= 6+4 = 10μC
V = Qt/C
Where Qt = Q1+Q2
V = Q1+Q2/C
V = 300+136/10
V = 437/10
V = 43.6volts
Uc1 = 1/2×C1V^2
= 1/2 × 6μF × 43.6^2
= 1/2 × 6μF × 1900.96
= 3μF × 1900.96volts
= 5702.88J
= 5702.88J/1000
= 5.7mJ
Answer:
5) 4.00, 2.00, 1.0
Explanation:
wave equation is given as;
F₀ = V / λ
Where;
F₀ is the fundamental frequency = first harmonic
Length of the string for first harmonic is given as;
L₀ = (¹/₂) λ
λ = 2 L₀
when L₀ = 1
λ = 2 x 1 = 2m
when L₀ = 2m
λ = 2 x 2 = 4m
For First harmonic, the wavelength is 2m, 4m
For second harmonic;
L₁ = (²/₂)λ
L₁ = λ
When L₁ = 1
λ = 1 m
when L₁ = 2
λ = 2 m
For second harmonic, the wavelength is 1m, 2m
Thus, the wavelength that could represent harmonics present on both strings is 4m, 2m, 1 m
Answer:
B. 4 m/s
Explanation:
v=d/t
Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.
