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Virty [35]
3 years ago
9

Initially stationary, a train has a constant acceleration of 0.8 m/s2. (a) What is its speed after 27 s? m/s (b) What is the tot

al time required for the train to reach a speed of 41 m/s?
Physics
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

(a) v1 = 21.6 m/s

(b) t = 51.25 s

Explanation:

Use kinematics equation

v1 = v0 + at

Given

v0 = 0 = initial velocity

a = 0.8 m/s^2 = acceleration

(a) t = 27 seconds

v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s

(b) v1 = 41 m/s

v1 = v0 + at

solve for t

t = (v1-v0)/a = (41-0)/0.8 = 51.25 s

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A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t
Marrrta [24]

Answer:

a) U = 652.545\,kJ, b) v \approx 659.568\,\frac{m}{s}

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

Q = U

The internal energy for a monoatomic ideal gas is:

U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T

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T = \frac{P\cdot V}{n\cdot R_{u}}

T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}

T = 52.325\,K

The thermal energy contained by the gas is:

U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)

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b) The physical model for the cat is constructed from Work-Energy Theorem:

U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

v = \sqrt{\frac{2 \cdot U}{m_{cat}}}

v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }

v \approx 659.568\,\frac{m}{s}

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