What is the rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod

Answers

Answer 1

Answer:

C) 5ML^2

Explanation:

2 Spheres of mass M

Bug's mass 3M

Rod length 2L, radius L

Find Rotational Inertia I

I=Σmr^2

I=(3M+M+M)L^2

I=5ML^2

Answer 2

The rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod is 5ML².

Rotational inertia of the system

The rotational inertia of the system is determined by summing all the masses together and multiply with the radius.

I(t) = Σmr²

Where;

m is the massr is the radius

I(t) = Inertia of the 2 spheres + inertia of the bug

I(t) = (M + M)L²  + (3M)L²

I(t) = 2ML² + 3ML²

I(t) = 5ML²

Thus, the rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod is 5ML².

Complete question is below:

There are 2 Spheres of mass M and Bug's mass 3M. The Rod length is 2L, radius L.

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Related Questions

What symbols are these?

Answers

Answer:

the bottom one is wollsiegel

how have astronomers interpreted the unexpectedly fast rotation of galaxies

Answers

Answer:

There must be a lot of dark matter that can be felt but not seen

The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.

What is a galaxy?

Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.

A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.

The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.

Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.

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Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
an unbalanced force of 42 Newtons has been applied?
A 75.60 kg
B 00.04 kg
C 23.33 kg
D 43.80 kg

Answers

Answer:

23.33 kg

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{42}{1.8} = 23.3333... \\ [/tex]

We have the final answer as

23.33 kg

Hope this helps you

Answer:

[tex]\boxed {\boxed {\sf C. \ 23.33 \ kg}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F=ma[/tex]

The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.

Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².

F= 42 kg*m/s² a= 1.8 m/s²

Substitute the values into the formula.

[tex]42 \ kg*m/s^1 = m * 1.8 \ m/s^2[/tex]

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}[/tex]

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m[/tex]

The units of meters per second squared cancel.

[tex]\frac {42 \ kg}{1.8 } =m[/tex]

[tex]23.3333333 \ kg=m[/tex]

Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.

[tex]23.33 \ kg \approx m[/tex]

The mass of the ball is approximately 23.33 kilograms.

Can u plz help me with my hw

Answers

Answer:

1) 4 x 25m = 100m

2) 0 because after 4 lengths, he's back at the starting block.

3) speed is distance over time so speed here is 100m/125s = 0.8m/s

4) ...

5) 100m / 1.25m/s = 80s

6.a) 100m / 0.5m/s = 200s

7.b) ... can't draw this now..

if you have 3 moles of iron, how many grams of iron do you have?

Answers

Answer:

55.84 look at periodic table

so 55.84*3= 167.52gFe

The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?

Answers

The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The  variation  of the volume is: ΔV = 7.67 cm³

Pressure is defined by the relationship between force and area.

       P = F / A

The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.

          PV = nR T

Where P is pressure, V is volume, and T is temperature.

Let's write this equation for two points assuming that the temperature has not changed.

          P₀ V₀ = P₁ V₁

          V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex]                 (1)

The subscript "o" is used for the start point and the subscript "1" for the end point.

The pressure in a barometer is:

         P = ρ g y

They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.

 

The volume of the cylinder is

         V = π r² y

Let's calculate the initial volume.

         V₀ = π 1 9

         V₀ = 28.27 cm³

We substitute in equation 1.

         V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]  

         V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]  

Let's calculate.

        V₁ = [tex]\frac{75}{59} \ 27.27[/tex]  

        V₁ = 35.94 cm³

The volume to be incremented is

         ΔV = V₁ - V₀

         ΔV = 35.94 - 28.27

         ΔV = 7.67 cm³

Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The change of the volume is: ΔV = 7.67 cm³

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An "energy bar" contains 26 ggof carbohydrates.How much energy is this in joules?

Answers

I’m it’s about 3.4878 energy in joules

given. force of 88N and an acceleration of 4 m/s 2 what is the mass?

Answers

Mass, m=22 kg
It is given that Force, F = 88N
Acceleration, a=4 m/s 2

State a hypothesis about your estimated daily water use.

Answers

Answer

Estimates vary, but, on average, each person uses about 80-100 gallons of water per day, for indoor home uses.

Explanation:

how do i do this question in science?

Answers

simple subtract and ummmmmm.....

A bob attached to a string of length L = 1.25 m, initially found at the equilibrium
position, is given an initial velocity v = 0.8 m/s. The maximum displacement angle,
Omax, is: (Take g=10 m/s)
17
3.2
13
10.2
6.4
Clear selection
A hoh attached to a string of length 2 m is displaced by an angle of 8º and then

Answers

The maximum displacement angle of the bob is 13⁰.

The given parameters;

Length of the pendulum, L = 1.25 mInitial velocity of the bob, v = 0.8 m/s

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]

The maximum displacement angle is calculated as follows;

[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]

Thus, the maximum displacement angle of the bob is 13⁰.

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I need help in question 7, a and b.

Answers

Answer:

The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)

Explanation:

when light enters water, it bends. what does the amount of bending depend on?

Answers

Answer:

the change of speed!

Explanation:

if light speeds up or slow down more it bend more

How fast is an object going if it travels from San Diego to Anaheim in 1.25 hours (hr)? The distance from San Diego to Anaheim is 93 miles (mi).

a) 74.4 mi/hr

b) 116.25 mi/hr

c) 0.013 mi/hr

d) 84.25 mi/hr

e) None of the answers

Answers

Answer:

74.4 mph

Explanation:

Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.

speed = mi/hr

speed = 93/1.25

speed = 74.4 mph

lonic Naming Practice
Write the name of the following chemical compound:
Na3Cl1

Answers

Answer:

sodium chloride. .....

Ionic naming practice: Na3Cl1= Sodium Chloride

. Quỹ đạo chuyển động của một vật là

Answers

Answer:

you are very kind Po lang

7. A bus covers a certain distance in 60 minutes if it runs at a speed of 60 km/hr.
What must be the speed of the bus in order to reduce the time of journey by 40
minutes?

Answers

Answer:

90 Km/h

Explanation:

60 mins is 1 hr

so in an hour bus covers 60 Km

so new speed:

d/t

(60km/40mins)*60mins/h

90Km/h

Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerza?

Answers

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

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You throw a can of cranberry sauce straight up in the air with an initial velocity of 12 m/s. What is the maximum height the baseball reaches above your head.

Answers

Answer:

Δx=7.35m

Explanation:

This is a free fall problem (g= -9.8 m/s²).

Displacement: ? <--- what you're trying to find (max height)

Initial Velocity: 12 m/s

Final Velocity: 0 m/s <-- since the can will land back in your hands

Time: Missing <-- not needed to solve

Acceleration: -9.8 m/s²

The equation that you will use is vf²=vi²+2aΔx

Now plug the information into the equation!

(0m/s)²=(12m/s)²+2(-9.8m/s²)Δx

Multiply the 2 with the acceleration (-9.8m/s²)

(0m/s)²=(12m/s)²-19.6m/s²Δx

Now you need to square the final and initial velocity

0m/s=144m/s-19.6m/s²Δx

Move 144 to the other side of the equal sign

-144m/s=-19.6m/s²Δx

Divide both sides by -19.6m/s²

7.346m=Δx

You can round it to the tenth or hundreth place, up to you!

For this example, I'll round to the hundredth...

Δx=7.35m

I would double check this answer or review it yourself to see if it's correct.

Good luck with your studies!

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