Answer:
5.3×10⁴ m/s
Explanation:
From the question,
Momentum = mass× velocity
M = mV................ Equation 1
Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane
make V the subject of the equation
V = M/m.................. Equation 2
Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg
Substitute these values into equation 2
V = 1.6×10⁹/3.0×10⁴
V = 5.3×10⁴ m/s
Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
Answer:7 cm/s
Explanation:
Given
Particle move along curve
As it reaches the (2,3) its y coordinate is increasing at 14 cm/s
Differentiating y w.r.t time
Now at (2,3)
