Answer:
L = a 1,929 10⁴ m
a = 0.1 mm = 0.1 10⁻³ m, L = 1,929 m
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
Let's use trigonometry to find the breast
tan θ = y / x
As the angles are very small
tan θ = sin θ/ cos θ = sin θ = y / x
We replace
a y / L = m λ
L = a y / m λ
The red light has a wavelength of Lam = 700 nm = 700 10⁻⁹ m, in the third pattern it is m = 3
L = a 4.05 10⁻² / (3 700 10⁻⁹)
L = a 1,929 10⁴ m
To give a specific value we must know the width of the slit, suppose a value of a = 0.1 mm = 0.1 10⁻³ m
L = 1,929 m
Answer:
in which standard you are, i am typing your answer till please reply.
Answer: If one bulb goes out the other bulbs stay lit.
If there is a break in one branch of the circuit, current can still flow through the other branches.
Explanation:
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s

t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **
Ans: The vertical distance = y = M/(ρA)
Explanation:Support the vertical distance = y
Object's density = M/(A*h) (since A*h = volume)
By applying the condition,
(M/(Ah))/ρ = y/h
M/(ρAh) = y/h
y = M/(ρA)