Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
<u>T₂ = 95.56°C</u>
The insulin levels lead to the cause of type 2 diabetes
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
Horizontal component of force(Normal reaction) :
Since, box is on the verge of slipping :
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
The total amount of energy stays the same, but throughout the ride, the kinetic energy and the potential energy change, still adding up to the same number. At the top of the ride it has potential energy, and as it goes down the potential energy decreases and the kinetic energy increases. When it’s at the bottom of the first drop it has maxed out its kinetic energy, and minimized its potential energy. Friction slows down the car, and pushes on the cart with a force that is equal and opposite to the force being exerted in the track. The reason the track keeps going is because though it exerts and equal and opposite force the momentum of the objects is different, allowing the car to continue moving, however friction will slow it down until eventually it comes to a stop.
