We know that speed is a scalar unit while velocity is a vector. The signs are mostly concerned with the magnitude that the car should travel in, the direction is not as important.
Hope I helped :)
Answer:
1.52 nm
Explanation:
Using the De Broglie wavelength equation,
λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s
So, λ = h/p
λ = h/mv
substituting the values of the variables into the equation, we have
λ = h/mv
λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)
λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)
λ = 0.1518 × 10⁻⁸ m
λ = 1.518 × 10⁻⁹ m
λ = 1.518 nm
λ ≅ 1.52 nm
It mimics the movement of the waves
<span>Condensation is the change of
the substance from liquid to solid phase. Example of this is the formation of
ice. Vaporization is the change of substance from liquid to gas phase. Example of
this is the boiling of water. Deposition is the change of a substance from gas
to solid phase. Example of this is the formation of ice on a winter day. Sublimation
is the change of a substance from solid to gas phase. Example of this is dry
ice. The answer is letter C.</span>
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
