Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He
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Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, = 926 kg/m³
density of the wood, = 974 kg/m³
density of water, = 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.
The apparent weight of a 1.1 g drop of water is 4.24084 N.
<h3>What is Apparent Weight?</h3>- According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
- By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.
Solution:
N = Speed of rotation = 1250 rpm
D = Diameter = 45 cm
r = Radius = 22.5 cm
M = Mass of drop = 1.1 g
Angular speed of the water =
Apparent weight is given by
= 4.24084 N
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Question:
The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?