What are significant figures?
2 answers:
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Answer:
a = 129.663
Explanation:
We know that:
T = Ia
Where T is the torque, I is the moment of inertia and a is the angular aceleration:
First, we will find the moment of inertia using the following equation:
I =
Where M is the mass and R is the radius of the disk. Replacing values, we get:
I =
I = 0.904 kg*m^2
Second, we will find the torque using the following equation:
T = ()*(R)
Where is the force on one side and
is the force on the other side. Replacing values, we get:
T = (162N-63N)(1.184m)
T = 117.216N*m
Finally, we replace T and I on the initial equation as:
T = Ia
117.216N = (0.904)(a)
Solving for a:
a = 129.663
The answer for the missing labeling as (i) and (ii) will be (i)thermometer and (ii)condenser.
<h3>What is a burner?</h3>A burner is a device intended to guarantee that the flame is stabilized.
By creating a proper flow field to cause the initial temperature rise, a The fuel/air combination is heated from within the flame to the ignition temperature.
When the flame or heat is created in fuel-burning or heat-generating equipment.
Wood, coal, kerosene, LPG, petrol, and diesel are a few of the often utilized fuels. In order to cook, heat, power vehicles, and create electricity, we utilize fuels.
The barrel, collar, air holes, gas intake, gas valve, and base are the different pieces of a Bunsen burner. Each component is essential to the process of creating a flame.
Hence, the answers for the missing labeling will be (i) thermometer and (ii) condenser.
To learn more about the burner refer;
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Answer:
<em>a) the final angular speed is 0.738 rad/s</em>
<em>b) the change in kinetic energy = 0.3 J</em>
Explanation:
the two 1 kg objects have a total mass of 2 x 1 = 2 kg
radius of rotation of the objects = 0.9 m
moment of inertial of the student and the chair = 6 kg-m^2
initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s
final angular speed of rotation of the sitting student and object system ω2 = ?
moment of inertia of the rotating object is
= 2 x
= 1.62 kg-m^2
total moment of inertia of sitting student and object system will be
==> 6 + 1.62 = 7.62 kg-m^2
The initial angular momentum of the sitting student and object system will be calculated from
==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2
if the radius of rotation of the object is reduced to 0.39 m,
new moment of inertia of the rotating object will be
= 2 x
= 0.304 kg-m^2
new total moment of inertia of the sitting student and object system will be
==> 6 + 0.304 = 6.304 kg-m^2
The final momentum of the sitting student and object system will be calculated from
==> Iω2 = 6.304 x ω2 = 6.304ω2
<em>According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system</em>. Therefore,
4.65 = 6.304ω2
ω2 = 4.65/6.30 =<em> 0.738 rad/s</em>
b) Rotational kinetic energy of the system =
for the initial conditions, kinetic energy is
==> =
= 1.417 J
for the final conditions, kinetic energy is
==> =
= 1.717 J
change in kinetic energy = final KE - initial KE
==> 1.717 - 1.417 = <em>0.3 J</em>