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Lapatulllka [165]
2 years ago
15

What is the average acceleration of a car that is initially at rest at a stoplight and then accelerates to 24 m/s in 9.4 s?

Physics
2 answers:
andrezito [222]2 years ago
8 0

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

topjm [15]2 years ago
5 0

Answer:

B) a= 2.6 m/s²

Explanation:

Conceptual analysis:

Because the car moves with uniformly accelerated movement, we apply the following formula :

vf= v₀+a*t Formula (1)

Where:    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Known Data

v₀=0

vf = 24 m/s

t=  9.4

Development problem

We replace the data in the formula (1)

vf= v₀+a*t

24 m/s = 0 + a*9.4s

a =  (24 m/s) / (9.4s)

a= 2.6 m/s²

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A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
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Answer:

Given that

V= 0.06 m³

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n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

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2 years ago
Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter
Mila [183]
The frequency of the radio wave is:
f=102.1 MHz = 102.1 \cdot 10^6 Hz

The wavelength of an electromagnetic wave is related to its frequency by the relationship
\lambda= \frac{c}{f}
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7 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
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Answer:

230

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Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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