A car engine burns gas at 495 K and exhausts to the air at 293 K. What is the highest possible efficiency it could have? (NO UNIT)

59.19%

Explanation:

Step 1:

Data obtained from the question. This include the following:

Input temperature = 495 K.

Output temperature = 293 K.

Efficiency =..?

Step 2:

Determination of the efficiency of the engine.

Efficiency is simply defined as the ratio of output to input times 100. Mathematically, it is expressed as:

Efficiency = output /input x 100

With the above formula, we can calculate the efficiency of the engine as follow:

Input temperature = 495 K.

Output temperature = 293 K.

Efficiency =..?

Efficiency = output /input x 100

Efficiency = 293/495 x 100

Efficiency = 59.19%

Therefore, the efficiency of the engine is 59.19%

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

there are no components of a physical goal mentioned on the list of choices that you've provided.