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2 years ago

Describe how the smoke particles move when a candle was blown out?

Physics

1 answer:

lukranit [14]2 years ago

4 0

Answer:

When a candle was blown out, the flame stops immediately but the wick and the wax are both still hot, so pyrolysis continues for a few seconds. Explanation:

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The law of conservation of mass states that mass can never be created or destroyed, only changed. true or faulse

Aleksandr-060686 [28]

Answer:

true

Explanation:

The law of conservation of mass states that in a chemical reaction mass is neither created nor destroyed. For example, the carbon atom in coal becomes carbon dioxide when it is burned. The carbon atom changes from a solid structure to a gas but its mass does not change.

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3 years ago

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

Deffense [45]

The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry

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3 years ago

One important relationship between chemical and mechanical weathering is ________. 1. mechanical weathering forms more brittle m

Paha777 [63]

Answer:

2. mechanical weathering can produce smaller pieces of rock that have more surface area for chemical weathering to work Explanation:

Mechanical weathering involves activities of living organisms or some geological processes. The bigger rocks are usually reduced to smaller rocks and further reduction might be limited or not posibble mechanically. This reduced rocks now increases the surface area available for chemical weathering; which further reduces the sizes of the rocks below the size range of mechanical weathering. one will recall that the rate of chemical reaction increases with exposed surface area.

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3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

3 years ago

In the 2016 Olympics in Rio, after the 50 mm freestyle competition, a problem with the pool was found. In lane 1 there was a gen

gogolik [260]

Answer:

Explanation:

still water speed is 50 m / 25.0 s = 2.00 m/s or 200 cm/s

In lane 1 the effective speed would be 201.2 cm/s

5000 cm / 201.2 cm/s = 24.85 s

The change is 25.00 - 24.85 = 0.15 s decrease in time

In lane 8, the effective speed would be 198.8 cm/s

5000 cm / 198.8 cm/s = 25.15 s

The change is 25.00 - 25.15 = 0.15 s increase in time

6 0

3 years ago

Describe how the smoke particles move when a candle was blown out?​

Describe how the smoke particles move when a candle was blown out?