Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.16, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking h = 0.0500 m for each. Figure 11.16 An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure P g = h ρ g transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure P g by an amount h ρ g .The jar's rigidity prevents atmospheric pressure from being transmitted to the peanuts.
Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.16, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking h = 0.0500 m for each. Figure 11.16 An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure P g = h ρ g transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure P g by an amount h ρ g .The jar's rigidity prevents atmospheric pressure from being transmitted to the peanuts.
Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.16, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking
h
=
0.0500
m for each.
Figure 11.16 An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure
P
g
=
h
ρ
g
transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure
P
g
by an amount
h
ρ
g
.The jar's rigidity prevents atmospheric pressure from being transmitted to the peanuts.
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 =
simple diagram to illustrate the setup for each law- coulombs law and biot savart law
Human Biology: Concepts and Current Issues (8th Edition)
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