Answer: 4.7rad
Explanation:
Angular displacement =s/r
Where s=distance traveled
r=radius
Angular displacement =141m/30m
Angular displacement =4.7rad.
You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".
Answer:
C = 771.35 J/kg°C
Explanation:
Here, e consider the conservation of energy equation. The conservation of energy principle states that:
Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container
Since,
Heat Given or Absorbed by a material = m C ΔT
Therefore,
m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃
where,
m₁ = Mass of Metal Piece = 2.3 kg
C = Specific Heat of Metal = ?
ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C
m₂ = Mass of Metal Container = 3.8 kg
ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C
m₃ = Mass of Water = 20 kg
C₃ = Specific Heat of Water = 4200 J/kg°C
ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C
Therefore,
(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)
C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J
C = 252000 J/326.7 kg°C
<u>C = 771.35 J/kg°C</u>
The tension on the wire is 52.02 N.
From the question, we have
Density of aluminum = 2700 kg/m3
Area,
A = πd²/4
A = π x (4.6 x 10⁻³)²/4
A = 1.66 x 10⁻⁵ m²
μ = Mass per unit length of the wire
μ = ρA
μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²
μ = 0.045 kg/m
Tension on the wire = √T/μ
34 = √T/0.045
34² = T/0.045
T = 52.02 N
The tension on the wire is 52.02 N.
Complete question:
The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.
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