Answer:
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph
<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph
<u>3. Addition:</u>
You add the corresponding components:
To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse:
Answer:
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
Explanation:
In this exercise you are asked to observe the change in velocity in a projectile launch.
If we assume that the friction force is small, the velocity in the x-axis must be constant
vₓ = v₀ₓ
Therefore, the arrow (red) that represents this movement must not change in magnitude.
In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
Answer:
5
Explanation:
electrons can be more than det
I would let someone more confident in this field to answer this question but that sounds like a Chemical reaction
Answer:
Broglie wavelength: electron 1.22 10⁻¹⁰ m
, proton 2.87 10⁻¹² m
, hydrogen atom 7.74 10⁻¹² m
Explanation:
The equation given by Broglie relates the momentum of a particle with its wavelength.
p = h /λ
In addition, kinetic energy is related to the amount of movement
E = ½ m v²
p = mv
E = ½ p² / m
p = √2mE
If we clear the first equation and replace we have left
λ = h / p =
λ = h / √2mE
Let's reduce the values that give us SI units
1 ev = 1,602 10⁻¹⁹ J
E1 = 100 eV (1.6 10⁻¹⁹ J / 1eV) = 1.6 10⁻¹⁷ J
We look in tables for the mass of the particle and the Planck constant
h = 6,626 10-34 Js
me = 9.1 10-31 Kg
mp = 1.67 10-27 Kg
Now let's replace and calculate the wavelengths
a) Electron
λ1 = 6.6 10⁻³⁴ / √(2 9.1 10⁻³¹ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 5.39 10⁻²⁴
λ1 = 1.22 10⁻¹⁰ m
b) Proton
λ2 = 6.6 10-34 / √(2 1.67 10⁻²⁷ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 2.3 10⁻²²
λ2 = 2.87 10⁻¹² m
c) Bohr's first orbit
En = 13.606 / n2 [eV]
n = 1
E1 = 13.606 eV
E1 = 13,606 ev (1.6 10⁻¹⁹ / 1eV) = 21.77 10⁻¹⁹ J
λ3 = 6.6 10⁻³⁴ /√(2 1.67 10⁻²⁷ 21.77 10⁻¹⁹) = 6.6 10⁻³⁴ / 8.52 10⁻²³
λ3 = 0.774 10⁻¹¹ m = 7.74 10⁻¹² m
