To find or discover by investigation?
Answer:
Minimum thickness will be 100 nm
Explanation:
We have given refractive index is n = 1.5
Wavelength of the light incidence
= 600 nm
We have to find the smallest thickness of the film so that there will be minimum light reflect
For minimum thickness of non reflecting film
, here t is thickness,
is wavelength and n is refractive index
Putting all values 
So minimum thickness will be 100 nm
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
Strong alien you got there good luck bud you never asked a question
Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest