Answer:
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thenks for the points :))
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.
<h3>What is the rate of heat transfer?</h3>
Rate of heat transfer is the power rating of the machine.
Work done and changes in potential and kinetic energy are neglected since it is a steady state process.
The specific heat in terms of specific heat capacity and temperature change is given as:
The rate of heat transfer, is then determined as follows:
- Qout = flow rate × specific heat
Qout = 0.75 × 20.08 = 15.06 kW
Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.
Learn more about rate of heat transfer at: brainly.com/question/17152804
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Answer:
Reflection
Explanation:
The specific behavior of light that will be essential to ensure the success of your design is "Reflection". This is because light maze makes use of a mirror and it's the light that is reflected that we see with our eyes. Also, the manner in which light is reflected off objects will affect the colors that are reflected as well.
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
So we equate the COP of our heater with COP of Carnot heater
Rearrange the equation
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C