Sam is traveling at 25 m/s for a total of 50 seconds how far does Sam Travel
1 answer:
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If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
Answer:
(a) 0.032 nm
(b) 39,235 eV
(c) 70,267.8 eV
Explanation:
(a) The energy of a photon can be calculated using:
E = hc/λ equation (1)
where:
h = 4.13*10^-15 eV.s
c = 3*10^8 m/s
λ = 0.024*10^-9 m
Thus:
E = (4.13*10^-15)*(3*10^8)/0.024*10^-9 = 51,625 eV
Then we calculate 76% of this estimated energy and determine the new wavelength:
Using equation (1) to determine the new wavelength:
λ
λ = (4.13*10^-15)*(3*10^8)/39235 = 3.15*10^-11 m = 0.032 nm
(b) As calculated in part (a), the maximum x-ray energy this machine can produce is
(c) The energy of a Ka x-ray photon can be estimated using:
where Z is the atomic number = 84.
= 70,267.8 eV