Leidenfrost effect . A water drop will last about 1 s on a hot skillet with a temperature between 100°C and about 200°C. However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance L in Fig. 18-48). Let L = 0.100 mm, and assume that the drop is flat with height h = 1.50 mm and bottom face area A = 4.00 × 10 –6 m 2 . Also assume that the skillet has a constant temperature T s = 300°C and the drop has a temperature of 100°C. Water has density ρ = 1000 kg/m 3 , and the supporting layer has thermal conductivity k = 0.026 W/mžK. (a) At what rate is energy conducted from the skillet to the drop though the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last? Figure 18-48 Problem 62.
Leidenfrost effect . A water drop will last about 1 s on a hot skillet with a temperature between 100°C and about 200°C. However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance L in Fig. 18-48). Let L = 0.100 mm, and assume that the drop is flat with height h = 1.50 mm and bottom face area A = 4.00 × 10 –6 m 2 . Also assume that the skillet has a constant temperature T s = 300°C and the drop has a temperature of 100°C. Water has density ρ = 1000 kg/m 3 , and the supporting layer has thermal conductivity k = 0.026 W/mžK. (a) At what rate is energy conducted from the skillet to the drop though the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last? Figure 18-48 Problem 62.
Leidenfrost effect. A water drop will last about 1 s on a hot skillet with a temperature between 100°C and about 200°C. However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance L in Fig. 18-48). Let L = 0.100 mm, and assume that the drop is flat with height h = 1.50 mm and bottom face area A = 4.00 × 10–6 m2. Also assume that the skillet has a constant temperature Ts = 300°C and the drop has a temperature of 100°C. Water has density ρ = 1000 kg/m3, and the supporting layer has thermal conductivity k = 0.026 W/mžK. (a) At what rate is energy conducted from the skillet to the drop though the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?
For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you
A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).
What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V)
ammeter
I =
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