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nlexa [21]
3 years ago
6

An airplane travels with a constant velocity of 210 m/s and in the upper atmosphere where the plane is traveling there is a wind

that is blowing at a constant velocity of 60 m/s to the east. Determine the resultant velocity for the plane when it is traveling
Chemistry
1 answer:
kolezko [41]3 years ago
4 0

Answer:

hello your question is incomplete the options are missing

Determine the resultant velocity for the plane when it is travelling

i) To the east

ii) To the west

answer :i)  270 i

             ii)  -150 i

Explanation:

velocity of Airplane = 210 m/s

wind velocity = 60 m/s to the east

The resultant velocity for the plane when it is travelling  

let the velocity of the wind = V2

           velocity of the plane = v1

i) The resultant velocity for the plane when travelling to the east

Vr = V2 i  + V1 i

Vr= 60i + 210i  = 270i

ii) resultant velocity when the plane is travelling to the west

Vr = - V1 i + V2i

     = -210i  + 60 i = -150 i  

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Silicon is the element having a mass of 28.09 g

<u>Explanation</u>:

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4 years ago
The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma
Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})}  \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}

By replacing the corresponding values we obtain:

\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8

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Best regards.

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3 years ago
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garik1379 [7]
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