Take the tiny bit of carbon dioxide and the tiny bit of water vapor out of the air,
and the rest of what you're breathing right now is a mixture of elements.
Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
Answer:
f = 55mm, h ’= -9.89 cm
f = 200 mm, h ’= 42.5 cm
Explanation:
For this exercise let's start by finding the distance to the image, using the equation of the constructor

where f is the focal length, p and q are the distances to the object and image, respectively
lens with f₁ = 55mm = 0.55cm
=
= 1.718
q₁ = 0.582 m
lens with f₂ = 200mm = 2m
=
= 0.4
q₂ = 2.5 m
the magnification of a lens is given by
m =
h ’=
let's calculate for each lens
f = 55mm
h '= - 0.582 / 10 1.7
h ’= 0.0989 m
h ’= -9.89 cm
f = 200 mm
h '= - 2.5 / 10 1.7
h ’= -0.425 m
h ’= 42.5 cm
The negative sign indicates that the image is real and inverted
Hi, what is it you need help with?
Answer:
3.16X10∧-11 m
Explanation:
1/2 mv2 = qV (KE = Electric potential energy)
velocity = √2qV/m = √( 2X 1.6X10∧-19 X 1500/9.11X10∧-31)
2.3X10∧7m/s
now use De Broglie equation
λ = h/mv
= 6.62X10∧-34/( 9.11X10∧-31 X 2.3X10∧7)
3.16 X 10∧-11 m
or
use the above equations and substitute to get the final eqiation
λ = h/√(2mqV) = 3.16X 10∧-11 m