Answer:
The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km
Explanation:
Given:
d = distance = 0.679 m
λ = wavelength of the light = 537 nm = 537x10⁻⁹m
dp = pupil diameter = 4.81 mm = 0.00481 m
Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?
For the separation of the peak from the central maximum it is:
In this case, the two small sources of the headlights have the same angle as the images that form inside the eye
Answer: The acelaration is 2 m/s^2
Explanation: Using the second Newton law
The total force acting over the object is equal to the mass x accelaration, that is:
We considerer the applied force to the object equal to 25N
and the friction force of 15 N
25N-15N= m*a
Answer:
(a) a = (2i + 4.5j) m/s^2
(b) r = ro + vot + (1/2)at^2
Explanation:
(a) The acceleration of the particle is given by:
vo: initial velocity = (3.00i -2.00j) m/s
v: final velocity = (9.00i + 7.00j) m/s
t = 3s
by replacing the values of the vectors and time you obtain:
![\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%3D%5Cfrac%7B1%7D%7B3s%7D%5B%289.00-3.00%29%5Chat%7Bi%7D%2B%287.00-%28-2.00%29%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7Ba%7D%3D%282%5Chat%7Bi%7D%2B4.5%5Chat%7Bj%7D%29m%2Fs%5E2)
(b) The position vector is given by:
where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2
It'd be A. Picture one shows a chemical change. Rust is made from the iron reacting with oxygen.
Answer:
a) # buses = 7
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
y =
t - ½ g t²
Let's decompose the speed, with trigonometry
v₀ₓ = v₀ cos θ
= v₀ sin θ
v₀ₓ = 40 cos 32
v₀ₓ = 33.9 m / s
= 40 sin32
= 21.2 m / s
When it arrives it is at the same initial height y = 0
0 = ( - ½ gt) t
That has two solutions
t = 0 when it comes out
t = 2 / g when it arrives
t = 2 21.2 /9.8
t = 4,326 s
We use the horizontal displacement equation
x = vox t
x = 33.9 4.326
x = 146.7 m
To find the number of buses we can use a direct proportions rule
# buses = 146.7 / 20
# buses = 7.3
# buses = 7
The distance of the seven buses is
L = 20 * 7 = 140 m
b) let's look for the scope for this jump
R = vo2 sin2T / g
R = 40 2 without 2 32 /9.8
R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)
