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3 years ago

suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.

1 answer:

7 0

We know that the mass of the log before burning is 5 kg. After it is burnt, the mass of the ash comes out to be 1 kg. So there is a definite mismatch of 4 kg. When the log is burnt, there is the formation of smoke. The air and water present in the wood vaporises to reduce the weight significantly. The combustion actually leads to the formation of carbon dioxide that mixes with the air and also water. The extra mass thus goes into the formation of carbon dioxide and water.

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If a much heavier stone rolled off the same cliff, would it hit the ground more quickly? explain

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No, the rate of gravity remains constant

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3 years ago

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The net reaction of the calvin cycle is the conversion of co2 into the three-carbon sugar g3p. Along the way, reactions rearrang

WINSTONCH [101]

The conversion of CO2 into the three-carbon sugar G3p is the Calvin cycle's net reaction.

<h3 /><h3>What is Calvin Cycle?</h3>

Calvin cycle uses ATP and NADPH, which are created during the light reaction, to decrease atmospheric carbon dioxide in 3C sugar; Three steps can be distinguished in the Calvin Cycle reaction:

Carboxylation: A reaction that is catalyzed by RUBISCO, where RUBIP is the CO2 acceptor, results in the formation of 6 molecules of PGA (phosphoglycerate).

Here, there are 6 molecules with 18 carbons in PGA and 3 molecules with 3 carbons in CO2.

Reduction: One molecule of the output, glyceraldehyde 3 phosphate, is used to create sugar while the remaining five molecules go through regeneration.

Here, one glyceraldehyde-3-phosphate molecule with three carbon atoms is present.

Regeneration: From glyceraldehyde 3 phosphate, CO2 acceptor (RUBIP) is generated.

Here, 3 molecules of RUBIP and 5 molecules of dihydroxy acetone phosphate each have 15 carbons.

Know more about the Calvin cycle here:

brainly.com/question/15205817

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1 year ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago

4) The components of Vector A are given as follows Ax = -2.4 Ay = + 3.8 What is the magnitude of the A, and what is the angle th

Monica [59]

Answer:

A = 4.49

α = 57.72°

Explanation:

Knowing the magnitude of x & y of a vector we can determine the total magnitude of a vector.

$A=\sqrt{A_{x}^{2} +A_{y}^{2} } \\A=\sqrt{(2.4)^{2} +(3.8)^{2} }\\A=4.49$

The angle tangent can be used to determine the angle.

$tan(\alpha )=\frac{3.8}{2.4}\\tan(\alpha ) =1.5833\\\alpha =tan^{-1}(1.5833) \\\alpha = 57.72 (deg)$

5 0

3 years ago

A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0

Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

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3 years ago