Answer;
-A wave with the longest wavelength.
Explanation;
-Diffraction is the apparent of wave through,around small obstacles and the spreading out of wave past small openings. When thinking of diffraction of a wave think of shining a flashlight around a corner. The light bends around the corner but there is a place where it is dark and the light does not hit. Diffraction of a wave is basically the wave bending around an object then dispersing out.
-The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. When the wavelength of the waves is smaller than the obstacle, no noticeable diffraction occurs.
Answer:
a = 8 m/s^2, Ffriction = 10 N, μk = 0.205
Explanation:
a. Force = Mass*Acceleration,
(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)
40 N = 5 kg*acceleration,
a = 40/5 = 8 m/s^2
b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.
Fapplied - Ffriction = m*a,
40 - Ffriction = 5*6,
40 - Ffriction = 30,
Ffriction = 40 - 30 = 10 N
c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".
10 = μk*Fnormal (Fnormal = m*g = 5*9.8)
10 = μk*49,
μk=10/49 ≈ 0.205
Answer:
At dawn your location on earth is pointed straight in the direction of the Earth's travel in its orbit. Between midnight and dawn you are moving head-on through the location of the meteors in space, which means that you will, on average, observe more of them.
- public.nrao.edu
Explanation:
hope this helps
Answer:
277.78 hours
Explanation:
The formula for calculating the amount of charge is expressed as;
Q = It
I is the current
t is the time
Given
I =0.05A
Q = 50,000C
Required
Time t
Recall that: Q = It
t = Q/I
t = 50,000/0.05
t = 1,000,000secs
Convert to hours
1,000,000secs = 1,000,000/3600
1,000,000secs = 277.78 hours
Hence it will take 277.78 hours for the charge to flow through the diode
