At first glance, this statement seems to be true. But after about a
microsecond of further consideration, one realizes that the statement
would actually set Boyle spinning in his grave, and is false.
Boyle's law states that there is a firm relationship among the pressure,
temperature, and volume of an ideal gas, and that you can't say anything
about how any two of these quantities depend on each other, unless you
also say what's happening to the third one at the same time.
As the pressure of an ideal gas increases, the volume will decrease in
direct proportion to the volume, IF THE TEMPERATURE OF THE GAS
REMAINS CONSTANT.
If you wanted to, you could increase the pressure AND the volume of an
ideal gas both at the same time. You would just need to warm it enough
while you squeeze it.
Answer:
eu não consigo entender essa situação o Fim dessa história só pode ter confusão dos iguais com cabeças diferentes a invés de amigas são concorrentes
Answer:
C 350W
Explanation:
Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s
m = 70kg and g =9.8m/s².
x = horizontal distance covered
Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.
E = mgh + 300t
Where t is the time taken to cover the distance
x = vt and h = 0.05vt
So
E = mg×0.05×vt + 300t
Substituting respective values
E = 70×9.8×0.05×1.4t +300t = 348t
P = E/t = 348W ≈ 350W.
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
