GO Icicles. Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig 18-55). Because the water–ice interface must have a temperature of 0°C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance L ) to the top of the icicle, where the temperature T r can be below 0°C. Take L = 0.12 m and T r = −5°C. Assume that the central tube and the upward conduction path both have cross-sectional area A. In terms of A , what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is 0.400 W/mžK, and the density of liquid water is 1000 kg/m 3 . Figure 18-55 Problem 78.
GO Icicles. Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig 18-55). Because the water–ice interface must have a temperature of 0°C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance L ) to the top of the icicle, where the temperature T r can be below 0°C. Take L = 0.12 m and T r = −5°C. Assume that the central tube and the upward conduction path both have cross-sectional area A. In terms of A , what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is 0.400 W/mžK, and the density of liquid water is 1000 kg/m 3 . Figure 18-55 Problem 78.
GOIcicles. Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig 18-55). Because the water–ice interface must have a temperature of 0°C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance L) to the top of the icicle, where the temperature Tr can be below 0°C. Take L = 0.12 m and Tr = −5°C. Assume that the central tube and the upward conduction path both have cross-sectional area A. In terms of A, what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is 0.400 W/mžK, and the density of liquid water is 1000 kg/m3.
You are standing a distance x = 1.75 m away from this mirror. The object you are looking at is y = 0.29 m from the mirror. The angle of incidence is θ = 30°. What is the exact distance from you to the image?
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 \((-180^{\circ},180^{\circ}]\).
Human Biology: Concepts and Current Issues (8th Edition)
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