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Aleks04 [339]
3 years ago
14

An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i

nitial velocity?
Physics
1 answer:
zmey [24]3 years ago
5 0

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

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Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on
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Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

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Read 2 more answers
At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?
photoshop1234 [79]
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2}  )=12.38
And so
\ln( \frac{K_2}{K_1})=12.38
And using K_1=6.1\cdot 10^{-8} s^{-1} , we find K2:
K_2=K_1 e^{12.38}=0.0145 s^{-1}


5 0
3 years ago
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