Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Wavelength- <span>distance between successive crests of a wave.
frequency- t</span><span>he rate at which something occurs or is repeated over time.
amplitude-</span><span> maximum extent of a vibration.</span>
Using Fleming’s left hand rule,
The magnetic force on the electron will be west. Example: If it is dropped above the equator in Africa, the electron will head towards North America.
Answer:
h f = W + KE
Input energy equals work function plus KE of emitted electron
W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19
W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)
W = (16.6 - 10.1)E-19 = 6.5E-19 J
h f = 6.5E-19 J for electrons to be emitted with zero KE
f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)
Uhm, i'm pretty sure it's deadly.