Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
Answer: a. E =9.9*EXP(-19)J
1 mole E= 596178J
b. E= 1.32*EXP(-15)J, 1 mole E=795MegaJ
c. E= 1.98*EXP(-23)J
1 mole E = 11.9J
Explanation: The Energy of a photon E, the wavelength are related by
E= h*c/wavelength
h is the Planck's constant 6.6*EXP(-34)J.s
c is speed of light 3*EXP(8)m/s
h*c=1.98*EXP(-25)
Now let's solve
a. E = h*c/wavelength
= h*c/(200*EXP(-9)m
=9.9*EXP(-19)J
1 mole of a photon contian 6.022*EXP(23)photons by advogadro
Now to get the energy of 1 mole of the photon we have
9.9*EXP(-19)*6.023*EXP(23)
=596178J
b. E=h*c/150*EXP(-12)m
=1.32*EXP(-15)J
1 mole will have
1.32*EXP(-15)*6.022*EXP(23)J
=795*EXP(6)J
c. E= h*c/1*EXP(-2)m
=1.98*EXP(-23)J
1 mole of the photon will have
1.98*EXP(-23)J *6.022*EXP(23)
= 11.9J.
You will notice that the longer the wavelength of the photon the lesser the Energy it as.
NOTE: EXP represent 10^
Answer:
A) The event horizon, singularity, and the chute located between the two.
