If a freely falling object were somehow equipped with a speedometer, its speedreading would increase each second by
a) about 15 m/s.
b) a rate that depends on its initial speed.
c) a variable amount.
d) about 5 m/s.
e) about 10 m/s.

Answer:

1)   I_ pendulum = 2.3159 kg m², 2)  I_pendulum = 24.683 kg m²

Explanation:

In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum

Sphere

They indicate the density of the sphere roh = 37800 kg / m³ and its radius

r = 5 cm = 0.05 m

we use the definition of density

               ρ = M / V

               M = ρ V

the volume of a sphere is

                V = 4/3 π r³

we substitute

              M = ρ 4/3 π r³

we calculate

              M = 37800  4/3 π 0.05³

              M = 19,792 kg

Bar

the density is ρ = 32800 kg / m³ and its dimensions are 1 m,

0.8 cm = 0.0008 m and 4cm = 0.04 m

The volume of the bar is

               V = l w h

              m = ρ l w h

we calculate

              m = 32800 (1   0.008   0.04)

              m = 10.496 kg

Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is

               r_cm = 1 / M (M r_sphere + m r_bar)

               M = 19,792 + 10,496 = 30,288 kg

               r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))

               r_cm = 1 / 30,288 (5,248 + 20,7816)

               r_cm = 0.859 m

This is the center of mass of the pendulum.

1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:

Sphere I = 2/5 M r2

Bar I = 1/12 m L2

parallel axes theorem

                  I = I_cm + m D²

where m is the mass of the body and D is the distance from the body to the axis of rotation

Sphere

      m = 19,792 ka

the distance D is

                D = 1.05 -0.85

                D = 0.2 m

we calculate

               I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168

               I_sphere = 0.811472 kg m²

Bar

m = 10.496 kg

distance D

             D = 0.85 - 0.5

             D = 0.35 m

              I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576

              I_bar = 1.5044 kg m²

The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts

              I_pendulum = I_sphere + I_bar

              I_pendulum = 0.811472 +1.5044

              I_ pendulum = 2.3159 kg m²

this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m

2) The moment is requested with respect to the pivot point at r = 0 m

Sphere

        D = 1.05 m

         I_sphere = 2/5 M r2 + M D2

        I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82

        I_sphere = 21.84 kg m²

Bar

D = 0.5 m

      I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624

      I_bar = 2,84266 kg m 2

The pendulum moment of inertia is

       I_pendulum = 21.84 +2.843

       I_pendulum = 24.683 kg m²

This moment of inertia is about the turning point at r = 0 m

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately will be

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

Answer:

a) Angular speed(w) = 2.02rad/sec

b) 73J ( It is Inelastic Collision)

Explanation:

Given:

Mass=45kg

Length on each side = 1.5m side which is hangs vertically from a frictionless pivot at the center of its upper edge.

We need to calculate

(a) What is the angular speed and

(b) To know why the angular momentum conserved but not the linear momentum

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

Final answer:

The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.

Explanation:

The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.

The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.

Learn more about Kinetic Energy and Work here:

brainly.com/question/32752469

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Answer:

its 45 over 6

Explanation:the answer is in  the question

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Explanation: