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Yuri [45]
3 years ago
9

Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

etic energy?
Physics
2 answers:
Anon25 [30]3 years ago
6 0
Velocity=3.4m/sec
Mass=30kg
so kinetic energy=1/2mv^2
=1/2×30×3.4×3.4
=15×3.4×3.4
=15×11.56
=173.4 kg m per second square
Oliga [24]3 years ago
5 0

Answer: 173.4 Joules

Explanation: Kinetic energy is the energy possessed by an object by virtue of its motion.

K.E=\frac{1}{2}\times mv^2

m= mass of object

v = velocity

K.E=\frac{1\times 30kg\times (3.4ms^{-1})^2}{2}=173.4kgm^2s^{-2}=173.4J

(1kgm^2s^{-2}=1Joule)

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Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

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Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

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Finally, T2 will be:

T_{2}=278.80 K

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