When travelling on a two lane highway driving 50 to 55 mph, you need a 10-12 second gap in oncoming traffic to pass safety. At 55 mph, you will travel over 800 feet in 10-12 seconds so will an oncoming vehicle. That means you need over 1600 feet which is about 1/3 of a mile to pass safety. It is harder to see and judge the speed of oncoming vehicles that are traveling 1/3 of a mile or more away from you.
Answer:
positive ion
Explanation:
due to the fact it loses electrons there is 2 more protons than electrons
Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2
Substitute g=10m/s^2, t=5s, d=125m
Answer:
61.85 ohm
Explanation:
L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm
Let ω0 be the resonant frequency.
ω0 = 2357 rad/s
ω = 2 x 2357 = 4714 rad/s
XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm
Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm
Impedance, Z = 
Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm
Thus, the impedance at double the resonant frequency is 61.85 ohm.
Answer:
5525 N/C
Explanation:
Magnitude of electric field ( E ) = 3500 N/c
Direction of electric field : positive X axis
point charge ( q ) = -9.0 * 10^-9
<u>Calculate the Magnitude of the net electric field at (a) x = -0.20 m</u>
Magnitude = 5525 N/C
Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2
= 81 / 0.04 = 2025 N/c
<em>Therefore the magnitude of the net electric field </em>
= 2025 + 3500
= 5525 N/C
