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nata0808 [166]
3 years ago
14

a disk initially at rest rotates with constant angular acceleration through an angel of 50 rad in 5 sec find the angular speed o

f the disk just after 3 sec. please step by step
Physics
1 answer:
lisov135 [29]3 years ago
3 0

Recall the formula,

∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²

where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:

50 rad = 1/2 <em>α</em> (5 s)²

<em>α</em> = (100 rad) / (25 s²)

<em>α</em> = 4 rad/s²

Now find the angular speed <em>ω </em>after 3 s using the formula,

<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>

<em>ω</em> = (4 rad/s²) (3 s)

<em>ω</em> = 12 rad/s

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C_3H_8+5O_2\rightarrow3CO_2+4H_2O

From the above chemical equation, it is clear that 1 mole of propane requires 5 moles of oxygen to undergo complete combustion as the result of this reaction, 3 moles of carbon-di-oxide and 4 moles of water molecules will be produced.

Therefore one-half mole of propane requires 2.5 moles of oxygen to undergo complete combustion. And in the reaction, 1.5 moles of carbon-di-oxide and 2 moles of water molecules are produced.

Therefore the answer is, 'One-half of propane combines with 2.5 moles of oxygen to produce 2 moles of water and 1.5 moles of carbon dioxide'

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1 year ago
A beaker is filled 60 cm high with this liquid. What is the total pressure at the bottom of the beaker? (1 atm =1.013 x 105 Pa)
Amiraneli [1.4K]

Answer:

Total pressure =1.01*10^5 Pa

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given data:

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P =  Atmospheric pressure + pressure due to water column

pressure due to water =\rho*g*h

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height of water column = 60 cm =0.60 m

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A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
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Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

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I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

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The linear speed is

v = (L/2)ω

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v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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