He travels 24 kilometers in total
50m/s
I divided the 200m by 4s (so 200/4) to get 50m per second
Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.
Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:
where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:
where in this problem:
is the initial pressure of the gas
is the initial absolute temperature of the gas
is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:
