ok well I've tried it a few times. I added the two amounts based on hint two but i guess that is wrong. Do i even incorporate gauge pressure?
I canceled the atm pressure out since it'll be the same at the start and finish.
So i took 499/27 = 46.2/?
So I had (1/46.2)*(499/27) = .400...
I've tried to take:
2.82 x 10 ^ 6 Pa * 499 cm ^ 3 / 27.0 C = 2.82 x 10 ^ 6 Pa * 46.2 cm ^ 3 / ?
the question mark would be the final temperature.
or am i way off with this? You guys got to understand that our teacher like doesn't teach us and i would learn way more if someone could give...
Yes...and i have tried that in relation to plugging in the values that i do know in order to find the final temperature but my answers keep saying wrong!
If the air has been compressed than we know that the temperature is greater. I don't really know what to do with the gauge pressure...
The final temperature that it is at...
here:
Hint 1. How to approach the problem
Use the ideal gas law to relate the initial pressure, temperature, and volume to their final values. Calculate the final temperature given the initial and final values in the introduction. Also, be very...
1. A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 cm^3 of air at atmospheric pressure (1.01 x 10 ^5 Pa) and a temperature of 27.0 C. At the end of the stroke, the air has been compressed to a volume of 46.2...
So:
Hint 2. Find the magnitude of the buoyant force
Find , the magnitude of the buoyant force.
Which i found to be: V * Pf * g
and i know mass * acceleration = tension but i don't know how to combine all these terms in order to get a correct answer.
1. A ball of mass mass b and volume V is lowered on a string into a fluid of density Pf. Assume that the object would sink to the bottom if it were not supported by the string. What is the tension in the string when the ball is fully submerged but not touching the bottom. Solve using...
sorry. I was confusing myself...
So ..."Write an expression for the force exerted on the tank's bottom by the water in the tank. Keep in mind that the water tank is located on Mars, so weights depend on the acceleration due to gravity on that particular planet."
Therefore i took the...
Ok so i got the answer on the downward force by the air on the outside air on teh bottom of the tank. I took 88000 x 2.30 = 2.02 x 10^5 N
But i still can't firgure out what the pressure would be on the bottom of the tank by the water in the tank...
I took density * gravity * height to get...