South it is the same direction the car is already moving
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?
C- Incentives it’s like a reward ~!
Answer:
, it will sink
Explanation:
The density of an object is given by

where
m is the mass of the object
V is its volume
For the body in the problem, we have
m = 4 kg = 4000 g

Therefore, its density is

And the object will sink in water, because its density is larger than that of water, which is
. (an object sinks when its density is larger than that of water, otherwise it floats).
For the answer to the question above, I assume that the question is two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3,?<span>two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3, is electrically neutral.
</span>From Gauss's law:
<span>Flux = ∫c E . dA = q/eo </span>
<span>Since this surface encloses all </span>
<span>charge, we can simplify: </span>
<span>Flux = (q1+q2+q3)/eo </span>
<span>Flux = </span>
<span>( (1*10^-6)+(-1.9*10^-6)+(0) )/(8.85*10^-12) = -101694.92 N·m2/C</span>