Answer: [B]: Earth .
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The solution would be like this for this specific problem:
Given:
diffraction grating
slits = 900 slits per centimeter
interference pattern that
is observed on a screen from the grating = 2.38m
maxima for two different
wavelengths = 3.40mm
slit separation .. d =
1/900cm = 1.11^-3cm = 1.111^-5 m <span>
Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
<span>As sinθ ≈ y/D y/D =
λ/d λ = yd / D </span>
∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]
<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>
Answer:
A: 1.962
B: 3.924
Explanation:
g = G *M /R^2
g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807
g = 9.807*5lbs/R^2 the average brick is about 5 pounds.
g = 9.807*5*10^2. I'm assuming the height is around ten feet to help you out.
with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.
( M = whatever the brick weighs it's not specified in the question)
(R = the distance from the ground or how high the scaffold is)
(hopefully you can just plug your numbers in there hope this helps)
Answer 1) : 62.5 km/hour is the average velocity of the train.
2) The final velocity of the car at the end of 75 m is 14.69 m/s
Explanation:
1) Displacement of the train = 100 km + 150 km = 250 km
Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours
Average velocity=
62.5 km/hour is the average velocity of the train.
2) The acceleration of the car, a= 1.2 
Distance covered by the car,s = 75 m
Initial velocity of the car ,
= 6 m/s
Final velocity of thre car ,
=?
Using third equation of motion:


The final velocity of the car at the end of 75 m is 14.69 m/s
Answer:
In the middle of a dry cell, is a rod made of carbon. Around the carbon rod is a chemical paste. At the same time, the carbon rod becomes positively charged. When this happens, electrical current flows out of the cell when a conductor is attached between the cell's positive and negative terminals.