Answer:
E = 2,575 eV
Explanation:
For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength
E = h f
c = λ f
Where the Planck constant has a value of 6.63 10⁻³⁴ J s
Let's replace
E = h c / λ
Let's calculate for wavelengths
λ = 4.83 10-7 m (blue)
E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷
E = 4.12 10-19 J
The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J
E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 2,575 eV
The saw will tip towards the heavy boy. they are centered on the main pivot point so weight dosnt matter but as they go out. mass and weight will determin if the saw moves or not
3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Let us first find out the radioactive constant of Phosphorus 32.
Radioactive constant, λ =
Here, is half life of phosphorus 32 = 14.3 days
λ =
The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,
m=m₀e⁻(λt)
=
=
=
=
=
= 0.125 mg
The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.