A spherical shell has an outside radius of 2.60 cm and an inside radius of a. The shell wall has uniform thickness and is made o
1 answer:
Answer:
The answer is below
Explanation:
a) The volume of a sphere is:
Volume = (4/3)πr³; where r is the radius of the shell.
Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:
Volume of spherical shell = cm³
But Density = mass / volume; Mass = density * volume.
Therefore, mass of spherical shell = density * volume
mass of spherical shell = *
cm³
Mass of liquid = volume of inner shell * density of liquid
Mass of liquid =
Total mass of sphere including contents = mass of spherical shell + mass of liquid
Total mass of sphere including contents (M) = *
+
=
Total mass of sphere including contents (M) = (346 - 14.5a³) grams
b) The mass is maximum when the value of a = 0
M = 346 - 14.5a³
M' = 43.5a² = 0
43.5a² = 0
a = 0
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Answer:
The minimum scale reading during the trip was 560N.
Explanation:
So, in order to solve this problem, we must first prepare a drawing that will represent the situation. (See picture attached)
In the drawing we have the three situations represented. When the elevator goes up, when it travels at a constant velocity and when it decelerates.
We can now do an analysis of each of the situations:
When elevator starts moving:
When the elevator starts moving, it will be accelerated upwards, so the sum of forces will be equal to the mass times the acceleration. In this case, we have two forces, Weight and the normal force, which is the force the floor of the elevator is making upon the scale. This is the reading the scale will give you.
ΣF=ma
-W+N=ma
N=ma+W
As you may see, in this case the weight is added to the force applied by the elevator to accelerate, so the scale reading will be maximum here.
When the elevator has constant velocity:
If the elevator has constant velocity, then its acceleration will be zero, so the sum of the forces will be equal to zero.
ΣF=0
-W+N=0
N=W
On this stage, the scale will return a reading of 800N, since it will not be accelerated.
When the elevator is decelerating:
On the final stage of the trip, the elevator will have a negative acceleration (it is decelerating) so the sum of the forces will be equal to the product between the mass and the acceleration, so we get:
ΣF=-ma
-W+N=-ma
N=W-ma
In this final stage, we can see that the elevator's force is being subtracted from the weight due to the loss of velocity. This is where the scale will give you a minimum reading, so we analyze this stage.
Weight is found by multiplying mass and the acceleration of gravity:
W=mg
so we can rewrite our equation as:
N=mg-ma
when factoring the mass we get:
N=m(g-a)
the mass can be found by dividing the weight into the acceleration of gravity:
W=mg
m=80kg
we know that in the middle of the trip, the elevator will travel 6m in 1s, so the constant velocity is:
this is the initial velocity for the final stage. The acceleration of the final stage can be found with the following formula:
the final velocity is zero since it goes to a stop, so the formula becomes:
when substituting we get:
When substituting values we can now find the scale's reading:
N=560N
So the minimum scale reading during the trip was 560N
