A force is applied to a 2.9 kg mass and pro-
1 answer:
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Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
The negative sign of acceleration means that the ship slows down its velocity in order to land.
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work: