Answer:
λ = 3 10⁻⁷ m, UV laser
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
let's use trigonometry
tan θ = y / L
as in this phenomenon the angles are small
tan θ = 
= sin θ
sin θ = y / L
we substitute
a y / L = m λ
let's apply this equation to the initial data
a 0.04 / L = 1 600 10⁻⁹
a / L = 1.5 10⁻⁵
now they tell us that we change the laser and we have y = 0.04 m for m = 2
a 0.04 / L = 2 λ
a / L = 50 λ
we solve the two expression is
1.5 10⁻⁵ = 50 λ
λ = 1.5 10⁻⁵ / 50
λ = 3 10⁻⁷ m
UV laser
Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is
<em>x</em> = (7.2 m/s) <em>t</em>
The object's height <em>y</em> at time <em>t</em> is
<em>y</em> = 9.4 m - 1/2 <em>gt</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is
<em>v</em> = -<em>gt</em>
(a) The object hits the ground when <em>y</em> = 0:
0 = 9.4 m - 1/2 <em>gt</em>²
<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)
<em>t</em> ≈ 1.92 s
at which time the object's vertical velocity is
<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s
(b) See part (a); it takes the object about 1.9 s to reach the ground.
(c) The object travels a horizontal distance of
<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m
The resistance of a given conductor depends on its electrical resistivity (
), its length(L) and its cross-sectional area (A), as follows:
In this case, we have 
, 
and 
. So, the total resistance of the wire with length of 138m is:
Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
