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victus00 [196]
3 years ago
10

How can I show that the sphere of radius R performs a simple harmonic movement. how can i set its reference point and make the f

ree body diagram.
I have the torque sum equation which is equal to the moment of inertia by angular acceleration

Physics
1 answer:
Marysya12 [62]3 years ago
5 0

Explanation:

Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod).  There are three forces on the pendulum:

Weight force mg at the center of the sphere,

Reaction force in the x direction at the pivot,

Reaction force in the y direction at the pivot.

Sum the torques about the pivot O.

∑τ = I d²θ/dt²

mg (L sin θ) = I d²θ/dt²

For small θ, sin θ ≈ θ.

mg L θ = I d²θ/dt²

Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.

If you wish, you can use parallel axis theorem to find the moment of inertia about O:

I = Icm + md²

I = ⅖ mr² + mL²

mg L θ = (⅖ mr² + mL²) d²θ/dt²

gL θ = (⅖ r² + L²) d²θ/dt²

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A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the g
levacccp [35]

The actual weight of the gas = apparent weight + weight.

The actual weight = W_{A} + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = W_{A}

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/m^{2}

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight =  W_{A} + W

Learn more about density here: brainly.com/question/406690

8 0
2 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
How fast is Barry Allen ( The Flash) in season 5 of the Flash?
Burka [1]

Answer:  The Flash, Allen's top speed is Mach 3.3, or 2,532 miles per hour.

Explanation:

8 0
3 years ago
Light from distant galaxies most likely shows a ...red shift, indicating that the universe is expandingblue shift, indicating th
ELEN [110]

Answer:

red shift, indicating that the universe is expanding

Explanation:

Doppler effect occurs when a source of a wave (e.g. light, or sound waves) moves relative to an observer; as a result of this relative motion, the wavelength of the wave appears lengthened/shortened to the observer. Two situations can occur:

- The source of the wave is moving towards the observer - in this case, the wavelength of the wave becomes shorter. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the blue-end of the spectrum (blue-shift)

- The source of the wave is moving away from the observer - in this case, the wavelength of the wave becomes longer. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the red-end of the spectrum (red-shift)

In our universe, we observe a red-shift for all the distant galaxies: this means that these galaxies are moving away from us, so this is an indication that the universe is expanding.

5 0
4 years ago
In the graph, which region shows nonuniform positive acceleration?
cricket20 [7]

Answer: A.AB

Explanation:

This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.

As we can see in the attached image, the graph can be divided in four segments:

OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.

AB: In this segment the acceleration is changing at a nonuniform rate, since  in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>

BC: In this segment the acceleration is changing at a nonuniform rate, since  in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.

CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.

From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:

Segment AB

3 0
3 years ago
Read 2 more answers
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