A train slows from 44 m/s to 19 m/s in 9 seconds. What is the train's acceleration? Show your work.
1 answer:
Answer:
a = 2.77 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 19 [m/s]
Vo = initial velocity = 44 [m/s]
a = acceleration [m/s²]
t = time = 9 [s]
Note: The negative sign in the above equation means that the train is decreasing its velocity.
19 = 44 - a*9
9*a = 44 -19
a = 2.77 [m/s²]
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Answer:
<em>Heat of the reservoir is 461.38 K or 188.1 °C</em>
<em>The heating load is 18.705 kW</em>
Explanation:
COP = 8.7
working temperature = 248 °C = 248 + 273.3 = 521.3 K
work power W = 2.15 kW
reservoir temperature = ?
heating load Q = ?
We know that
COP = Q/W
Q = COP x W = 8.7 x 2.15 = <em>18.705 kW</em>
Also,
COP = =
8.7 =
4535.31 - 8.7 = 521.3
4535.31 - 521.3 = 8.7
4014.01 = 8.7
= 4014.01/8.7 = <em>461.38 K</em>
or 461.38 -273.3 = <em>188.1 °C</em>