<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>
Answer:
we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.
Explanation:
To design the experiment of measuring the critical angle, we describe the phenomenon, when the light passes from a medium with a higher refractive index to one with a lower index, it separates from the normal one and the Critical Angle is defined as the Angle for which the refraction occurs at 90º
n₂ sin θ₂ = n₁ sin 90
n₁ / n₂ = sin θ₂
As we can see, we have to measure the angle with which the laser touches the exit surface of the glass block.
Design of the experiment:
We place the glass block on the ramp and at the top we hit the conveyor for half the angle, we climb the block on the ramp and see that the angle of incidence of lightning on the exit face changes, part of the beam comes out of the glass , we see it by dispersion in the particles of dirty in the air; Maybe the conveyor or the laser should be moved slightly so that the beam touches the point of origin on the conveyor.
When we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.
Answer:
0.47 N
Explanation:
Here we have a ball in motion along a circular track.
For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.
This force is called centripetal force, and its magnitude is given by:
where
m is the mass of the object
is the angular velocity
r is the radius of the circle
For the ball in this problem we have:
m = 40 g = 0.04 kg is the mass of the ball
is the angular velocity
r = 30 cm = 0.30 m is the radius of the circle
Substituting, we find the force:
