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harkovskaia [24]
3 years ago
10

A solid disk of mass m1 = 9.3 kg and radius R = 0.22 m is rotating with a constant angular velocity of ω = 31 rad/s. A thin rect

angular rod with mass m2 = 3.9 kg and length L = 2R = 0.44 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.
1) What is the initial angular momentum of the rod and disk system?
6.97 kg-m2/s
2) What is the initial rotational energy of the rod and disk system?
108.14 J

3) What is the final angular velocity of the disk?

4) What is the final angular momentum of the rod and disk system?

5) What is the final rotational energy of the rod and disk system?

6) The rod took t = 6.5 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?
Physics
1 answer:
mel-nik [20]3 years ago
5 0
We are given with
<span>m1 = 9.3 kg
R = 0.22 m
ω = 31 rad/s
m2 = 3.9 kg
L = 2R = 0.44 m

The angular momentum is determined using the formula
L  = I w
where I is the moment of inertia

For a solid disk
I = (1/4) MR^2
For a rectangular rod
I = (1/12) ML^2

The rotational energy has the formula
E = (1/2) I w^2

Substitute the given values for getting the answers
</span>
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100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m
kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}

Put the value into the formula

f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}

f_{e}=3.68\ cm

Hence, The focal length of eyepiece is 3.68 cm.

3 0
3 years ago
Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0
3 years ago
What is the proportionality between pressure and temperature, and the proportionality between atmospheric pressure and tempratur
Alborosie

According to Ideal gasTo solve this problem, the fastest relationship allows us to observe the proportionality between the two variables would be the one expressed in the ideal gas equation, which is

PV= NRT

Here

P = Pressure

V = Volume

N = Number of moles

R = Gas constant

T = Temperature

We can see that the pressure is proportional to the temperature, then

P \propto T

This relationship can be extrapolated to all the scenarios in which these two variables are related. As the pressure increases the temperature increases. The same goes for the pressure in the atmosphere, for which an increase in this will generate an increase in temperature. This variable can be observed in areas of different altitude. At higher altitude lower atmospheric pressure and lower temperature.

6 0
2 years ago
when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0
3 years ago
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