F~1/r²
doubling the distance r, Decreases the force by ¼
Given:
Lens.........diameter ...fl#
eyepiece...2cm............5
objective...40cm........15
focal length of eyepiece = 2*5 = 10cm
focal length of objective = 40*15 = 600cm
magnification = FL obj / FL eyp = 600/10 = 60x
<span>step 1: energy required to heat coffee
E = m Cp dT
E = energy to heat coffee
m = mass coffee = 225 mL x (0.997 g / mL) = 224g
Cp = heat capacity of coffee = 4.184 J / gK
dT = change in temp of coffee = 62.0 - 25.0 C = 37.0 C
E = (224 g) x (4.184 J / gK) x (37.0 C) = 3.46x10^4 J
step2: find energy of a single photon of the radiation
E = hc / λ
E = energy of the photon
h = planck's constant = 6.626x10^-34 J s
c = speed of light = 3.00x10^8 m/s
λ = wavelength = 11.2 cm = 11.2 cm x (1m / 100 cm) = 0.112 m
E = (6.626x10^-34 J s) x (3.00x10^8 m/s) / (0.112 m) = 1.77x10^-16 J
step3: Number of photons
3.46x10^4 J x ( 1 photon / 1.77x10^-16 J) = 1.95x10^20 photons</span>
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
