Answer:
The heat absorbed in each cycle is 9,234.286 J
Explanation:
Given;
power output, P = 5 kW = 5,000 W
efficiency of the engine, e = 30 % = 0.3
thermal heat expelled, [tex]Q_c[/tex] = 6464 J
let the heat absorbed = [tex]Q_h[/tex]
The efficiency of the engine is given as;
[tex]e = \frac{W}{Q_h} = \frac{Q_h-Q_c}{Q_h} = \frac{Q_h}{Q_h} - \frac{Q_c}{Q_h} = 1-\frac{Q_c}{Q_h}\\\\e = 1-\frac{Q_c}{Q_h}\\\\0.3 = 1-\frac{Q_c}{Q_h}\\\\\frac{Q_c}{Q_h} = 1-0.3\\\\\frac{Q_c}{Q_h} = 0.7\\\\Q_h = \frac{Q_c}{0.7} \\\\Q_h = \frac{6464}{0.7} = 9,234.286 \ J.[/tex]
Therefore, the heat absorbed in each cycle is 9,234.286 J.
Please help meeeee
For a velocity versus time graph how do you know what the velocity is at a certain time? How do you know the acceleration at a certain time?
Answer:
Explained below
Explanation:
For a velocity time graph, the y - axis will represent velocity while the x - axis will represent time.
Now, to calculate velocity at a certain time t, we will draw a perpendicular line from the time on the x-axis to the graph line and trace the horizontal line from that point to the y-axis which will give the corresponding velocity at that time.
Now, for the acceleration at a time t. After getting the velocity like explained above, we now divide the velocity by the time.
How does light move?
Answer:
Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. It speeds through the vacuum of space at 186,400 miles (300,000 km) per second.
Explanation:
Hope this helps :))
Mary and John are seated on a bench a distance 2 part. Suddenly, John moves closer to Mary and claims that he has been drawn by the gravitational force. Is John’s claim justifiable? Show quantitatively. (Assume the masses of John and Mary are 67 and 55 respectively, and the coefficient of static friction between John and the bench is 0.15). [4]
Answer:
no
gravitational force between them is far smaller than the frictional force between the guy and the bench
A swimmer speeds up from 1.1 m/s to 3.2 m/s during the last 13.0 seconds of the race. What is the acceleration of the swimmer?
Answer:
a = 0.16 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t\\[/tex]
where:
Vf = final velocity = 3.2 [m/s]
Vo = initial velocity = 1.1 [m/s]
t = time = 13 [s]
a = acceleration [m/s²]
Now replacing:
[tex]3.2=1.1+a*13\\2.1=13*a\\a=0.16[m/s^{2} ][/tex]