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Julli [10]
3 years ago
9

The car wit fastest acceleration is the tempest. it has an acceleration of 6.1 m/s. suppose the car accelerates from rest to a f

inal speed of 30.m/s. how long does it take the tempest to reach this speed?
Chemistry
1 answer:
ziro4ka [17]3 years ago
3 0

<u>Given:</u>

Initial velocity (v1) = 0 m/s

Final velocity (v2) = 30 m/s

Acceleration (a) = 6.1 m/s2

<u>To determine:</u>

The time (t) taken to reach the final speed

<u>Explanation:</u>

Use the relation:

Acceleration (a) = [final velocity(v2) - initial velocity (v1)]/time (t)

t = (v2-v1)/a = 30-0/6.1 = 4.92 s

Ans: Time taken is around 4.9 s

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The irreversible elementary gas-phase reaction is carried out isothermally at 305 K in a packed-bed reactor with 100 kg of catal
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Answer:

0.856.

Explanation:

Lets represent the irreversible elementary gas phase equation of reaction as

A + B -----------------------------------> C + D

We have that the percentage of conversion is 80%.

The pressure, p from the ratio of exit pressure and entering pressure is p = 2/20 = 1/10 = 0.1.

Therefore, n = 1 - p^2/ weight of the catalyst = 1 - 0.1^2/ 100 = 9.9 × 10^-3 kg cat^-.

Now, let's make use of the equation below;

J/ 1 - J = kb^2/ u [ w - nw^2/2] ----------(1).

0.8 / 1- 0.8 = k ( 0.4)^2/ 10 [ 100 - (9.9× 10^-3 × 100^2/ 2] .

k = 4.95 dm^6/ kg.cat .mol.min

The turbulent flow= 1/2 × 9.9 × 10^-3 = 4.95 × 10^-3 kg cat^-.

Thus, making use of the equation (1) again, we have that;

{4.95 × 10^-3 × 0.4}/ 10 × [ 100 - (4.95 × 10^-3 × 100^2)] / 2 = 5.964.

Therefore, a/1 - a = 5.964.

5.964( 1 - a) = a.

5.964 - 5.964a = a.

5.964 = a + 5.964a.

5.964 = 6.964a.

a = 5.964/ 6.964 = 0.856.

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On January 28, 1986, the NASA shuttle orbiter mission STS-51-L and the tenth flight of Space Shuttle Challenger (OV-99) broke apart 73 seconds into its flight, killing all seven crew members, which consisted of five NASA astronauts and two payload specialists.

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motikmotik

Answer:

\boxed {\boxed {\sf 5.1 \ mol \ H_2O}}

Explanation:

To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.

We can use it in a ratio.

\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Multiply by the given number of molecules.

3.1*10^{24} \ molecules \ H_2O*\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Flip the ratio so the molecules of water cancel out.

3.1*10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O}{6.02*10^{23} \ molecules \ H_2O}

3.1*10^{24} *\frac {1 \ mol \ H_2O}{6.02*10^{23} }

\frac {3.1*10^{24} \ mol \ H_2O}{6.02*10^{23} }

Divide.

5.14950166113 \ mol \ H_2O

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

5.1 \ mol \ H_2O

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.

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C

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