Answer:
f1/f2 =W1/W2 = 1/3
.0 f2 = 3f1
As ,
1/F= 1/f1 +1/f2
...1/40 = 1/f1 - 1/3f1
f1=> 80/3 cm
... f2 = 2f1 = 3 x 80/3 = 80 cm
Y - yo = Vo*t - g * (t^2) / 2
Vo = - 9.0 m/s
t = 0.50 s
=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.
Answer: option c) - 5.7
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Answer:
<u><em></em></u>
- <u><em>1,500 kg.m/s</em></u>
Explanation:
First, arrange the information in a table:
Object Mass (kg) Velocity (m/s)
A 200 15
B 150 - 10
After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.
According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.
<u>Momentum before the collision, P₁</u>:
<u>Momentum after the collision</u>:
- As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
