A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency higher than that actually being produced by the object.
Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ + t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:
We have the output force (3000) and the mechanical advantage is 15, the formula is Ma = F(o) / F(i), so 15 = 3000 / x, so x = 200 newtons!
Answer:
Explanation:
a)
d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m
λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m
n = order of the bright fringe
D = screen distance = 130 cm = 1.30 m
= Position of nth bright fringe
Position of nth bright fringe is given as
For n = 1
For n = 2
For n = 3
b)
Position of nth dark fringe is given as
For n = 1
For n = 2
For n = 3
Answer:
x = 625 m
Explanation:
To solve this exercise we use the kinematics relations
v² = v₀² - 2 a x
indicate that the train stops therefore its final speed is zero (v = 0)
0 = v₀² - 2a x
x =
let's slow down to the SI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600 s) = 25 m / s
let's calculate
x =
x = 625 m