Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
Answer:
When a positive charged object is placed near a conductor electrons are attracted the the object. ... When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side. As you know, electrons are always moving. They spin very quickly around the nucleus of an atom. As the electrons zip around, they can move in any direction, as long as they stay in their shell.
Think of it like this, gravity has to pull harder on the heavier object to make them fall at the same rate , but doesn't have to pull as hard for the lighter object , thus is why sometimes heavier objects fall faster then lighter ones
Answer:
Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A
Explanation:
Em = 223 V
f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF
a)
Capacitive reactance = Xc=?
Xc= 
Xc=1/2pi *399*23.1*10^-6
Xc= 17.267 Ω
b).
Z=
Xl= 2π * f * L
Xl= 2π * 399 * 147 * 
Xl= 368.5 Ω
Z= = 
Z= 415.5 Ω
c).
Current:
I= V / Z= Em / Z
I= 223/415.5
I= 0.537 A
