Answer:
-6.44 m/s²
Explanation:
Given:
Δx = 60 m
v₀ = 27.8 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (27.8 m/s)² + 2a (60 m)
a = -6.44 m/s²
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
Answer:
a) 16 N
b) 2.13 m/s²
Explanation:
Draw a free body diagram of the tv stand. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force P pulling right.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
The net force in the x direction is:
∑F = P − Nμ
∑F = P − mgμ
∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)
∑F = 16 N
Net force equals mass times acceleration:
∑F = ma
16 N = (7.5 kg) a
a = 2.13 m/s²