Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.
Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.
After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.
Answer:
The student is getting different info bc the students probable keeping track of the distance instead of the displacement.
Explanation:
I would believe it to be C. Gold, but I'm not quite sure
Answer:
a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV
Explanation:
a. KE =1/2 (MV^2) where the M is mass of electron
b. E = V/d
c. V= 0 V (momentarily the pd changes to zero)
d KE= 300*3600 v = 1.08 MeV
