Answer:
Highest order: m = 1
Explanation:
Formula for solving this is;
dsin θ_m = mλ
We want to find m, thus;
m = (dsin θ_m)/λ
We are told that White light spans from 380nm - 750 nm. Thus, at maximum, λ = 750 nm.
θ_m = 90°
d = (900 lines/mm) = 9000 × 10^(-7) lines/nm
Since we want to find m, the units nm has to cancel out in the equation.
Thus, we will write d in nm/lines as (10^(7))/9000 nm/lines
Thus;
m = ((10^(7))/9000 × sin 90)/750
m = 1.48
Approximately 1
Answer:
a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²
Explanation:
a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force
F = m a
a = v² / r
F = G m M / r²
G m M / r² = m v² / r
G M / r = v²
The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships
v = d / t
As the orbit is circular the distance is the length of the circle in 24 h time
d = 2π r
t = 24 h (3600 s / 1 h) = 86400 s
Let's replace
G M / r = (2π r / t)²
G M = 4 π² r³ / t²
r = ∛(G M t² / (4π²)
r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)
r = 4.22 10⁷ m
b) the speed module is
v = √G M / r
v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷
v = 3.07 10³ m / s
c) the acceleration is
a = G M / r²
a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²
a = 0.224 m / s²
Answer:
m_{p} = 0.3506 kg
Explanation:
For this exercise we use Newton's equilibrium equation
B - Wc-Wp = 0
where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block
B = Wc + Wp
the state push for the Archimeas equation
B = ρ_water g V
the volume of the water is the area of the block times the submerged height h, which is
h´ = 8 - h
where h is the height out of the water
ρ_water g A h´ = 
g + 
g
ρ_water A h´ = m_{c} + m_{p}
write this equation to make the graph
h´= 1 /ρ_water A (m_{c} +m_{p})
h´ = 1 /ρ_waterA (m_{c} + m_{p})
if we graph this expression, we get an equation of the line
y = m x + b
where
y = h´
m = 1 /ρ_water A
b = mp /ρ_water A
whereby
m_{p} = b ρ_water A
ρ_water = 1000 kg / m³
b = 0.0312 m
m = 0.0890 m / kg
we substitute the slope equation
b = m_{p} / m
calculate
m_{p}= 0.0312 / 0.0890
m_{p} = 0.3506 kg
