Answer:
Whats the question/word problem or where is the graph (if included) representing this problem?
Answer:
1) λ = 0.413 m
, 2)v = 25,213 m / s
, 3) T = 0.216 N
, 4) m = 22.04 10-3 kg
Explanation:
1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related
λ = 2L / n n = 1, 2, 3 ...
In this case L = 0.62 m and n = 3
Let's calculate
λ = 2 0.62 / 3
λ = 0.413 m
2) the velocity related to wavelength and frequency
v = λ f
v = 0.413 61
v = 25,213 m / s
3) let's use the equation
v = √T /μ
T = v² μ
T = 25,213² 3.4 10⁻⁴
T = 0.216 N
4) the rope tension is proportional to the hanging weight
T-W = 0
T = W
W = m g
m = W / g
m = 0.216 / 9.8
m = 22.04 10-3 kg
5) n = 2
λ = 2 0.62 / 2
λ = 0.62 m
6) v = λ f
v = 0.62 61
v = 37.82 m / s
7) T = v² μ
T = 37.82² 3.4 10⁻⁴
T = 0.486 N
8) m = W / g
m = 0.486 / 9.8
m = 49.62 10⁻³ kg
9) n = 1
λ = 2 0.62
λ = 1.24 m
v = 1.24 61
v = 75.64 m / s
T = v² miu
T = 75.64² 3.4 10⁻⁴
T = 2.572 10⁻² N
m = 2.572 10⁻² / 9.8
m = 262.4 10⁻³ kg
For conservation of energy we have to:
mgH=mv²/2
Clearing
<span> v=sqrt(2gH)
Then, by definition
</span><span> F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt=
</span> =m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt
the answer is
F=m[sqrt(2gH)]/ Δt
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
