Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m
Answer:
The process of evaporative cooling helps the body is explained below in complete details.
Explanation:
Your body performs the application of the evaporative process when secreting. Sweat, which contains 90 percent water, commences evaporating. ... This appears in a cooling impression (described as evaporative cooling) that serves to sustain body temperature and cools the body down when it becomes too hot.
Answer:
the last one
Explanation:
because it is composed of atoms of two or more elements and carbon is a pure substance composed of only one type of atoms
Answer:
A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.
B. Veloctiy (Vb) = 1.66m/s
Explanation:
Given the following data
x(a) = 0.3m
x(b) = 0
q = 1.6×10^-19
Q = 24nc
r = 0.15m
Required: the motion of the electron and the velocity (Vb)
1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B
2. Potential energy and kinetic energy are given by
U(a) + K(a) = U(b) + K(b). . .1
Initial P.E and K.E are given as
U(a) = kQ/√x²(a) + a2
By substitution, we have
U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)
U(a) = -1.03×10^-16
Final P.E and K.E are given as
U(b) = KQ/√x²(b) + a2
By substitution, we have
U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²
U(b) = -2.3×10^16
3. By substitution into equation 1 becomes
-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2
V(b) = √2×1.27×10^-16/9.1×10^31
V(b) = 1.66×10^7m/s
The desk is in equilbrium, so Newton's second law gives
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
==> <em>n</em> = <em>mg</em>
==> <em>p</em> = <em>f</em> = <em>µn</em> = <em>µmg</em> = 0.400 (80.0 kg) <em>g</em> = 313.6 N
The student pushes the desk 3.00 m, so she performs
<em>W</em> = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J
of work.
