An airplane is moving horizontally with a constant velocity of (+ 115 m/s) at altitude of 1050 m. The directions to the right an
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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?
A) 19.6 m/s (downward)
B) 576 J
C) 19.6 m
D) Velocity: not affected, kinetic energy: doubles, distance: not affected
Explanation:
A)
An object in free fall is acted upon one force only, which is the force of gravity.
Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:
where:
v is the velocity at time t
u is the initial velocity
is the acceleration due to gravity
For the object in this problem, taking downward as positive direction, we have:
(the object starts from rest)
Therefore, the velocity after
t = 2 s
is:
(downward)
B)
The kinetic energy of an object is the energy possessed by the object due to its motion.
It can be calculated using the equation:
where
m is the mass of the object
v is the speed of the object
For the object in the problem, at t = 2 s, we have:
m = 3 kg (mass of the object)
v = 19.6 m/s (speed of the object)
Therefore, its kinetic energy is:
C)
In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
For the object in free fall in this problem, we have:
u = 0 (it starts from rest)
(acceleration of gravity)
t = 2 s (time)
Therefore, the distance covered is
D)
Here the mass of the object has been doubled, so now it is
M = 6 kg
For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.
For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.
The new kinetic energy is
where
M = 6 kg is the new mass
v = 19.6 m/s is the speed
Substituting,
And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.
Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.