Answer:
The final pressure of the gas is 9.94 atm.
Explanation:
Given that,
Weight of argon = 0.16 mol
Initial volume = 70 cm³
Angle = 30°C
Final volume = 400 cm³
We need to calculate the initial pressure of gas
Using equation of ideal gas
Where, P = pressure
R = gas constant
T = temperature
Put the value in the equation
We need to calculate the final temperature
Using relation pressure and volume
Hence, The final pressure of the gas is 9.94 atm.
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Use the equation I=V/R where I is current and V is the voltage plus R is the resistance so when voltage is the highest and resistance is lowest the current is the highest
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
