Devora tested two engines. as each engine's rotation got faster, its temperature increased at a constant rate.
the first engine's temperature (in degrees celsius) as a function of its rotation speed (in cycles per second) is given by the following table of values:
the graph of the second engine's temperature (in degrees celsius) as a function of its rotation speed (in cycles per second) is shown below.
the temperature of which engine increased more for each increase of 1 cycle per second?
choose 1

(choice a)
a
the first engine

(choice b)
b
the second engine

(choice c)
c
the temperatures of both engines increased the same
which engine was hotter when it rotated at 20 cycles per second?
choose 1

(choice a)
a
the first engine

(choice b)
b
the second engine

(choice c)
c
both engines were equally hot

The temperature of the second engine increases more for each increase of 1 cycle per second.

The second engine will be hotter.

Step-by-step explanation:

Each engine's rotation got faster, its temperature increased at a constant rate.  

The first engine's temperature (in degrees Celsius) as a function of its rotation speed (in cycles per second) is given by the table attached.

So, as the rate of change of temperature with increase in speed of rotation is linear, so, we can express the relation as  

T = aS + b (1)

Now, from the table we can write 26.4 = 8a + b (2)

and 28.8 = 11a + b (3)

Now, from equation (2) and (3), we get

11a - 8a = 2.4

⇒ 3a = 2.4  

⇒ a = 0.8  

Again, from equation (2) we get,

26.4 = 8 × 0.8 + b

⇒ b = 20

Therefore, the equation (1) becomes T = 0.8S + 20 (4)

Now, from the graph of the second engine, the y-intercept is 22 and slope of the straight line is {Since, the graph passes through the points (0,22) and (5,28).

Therefore, the equation can be written as  

T = 1.2S + 22 (5)

Therefore, the temperature of the second engine increases more for each increase of 1 cycle per second as from equations (4) and (5) the rate of change of temperature with respect to rotational speed is 1.2 for the second engine which is more than 0.8 for first engine. (Answer)

Now, for S = 20 cycles per second, from equation (4) we get,

T = 16 + 20 = 36° C

And from equation (5) we get, T = 24 + 22 = 46° C

So, the second engine will be hotter. (Answer)

step-by-step explanation:

3024 rhe base is what up in worry about