That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:
The displacement is the distance and direction from the place
where the crow took off to the place where he landed.
-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.
D² = (60 km)² + (45 km)²
= 3,600 km² + 2,025 km² = 5,625 km²
D = √(5625 km²) = 75 km .
-- It's direction is the angle whose tangent is (45 S / 60 W).
tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west
= 53.1° west of south.
= not exactly southwest but close.
Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
Answer:
The fundamental frequency of can is 2.7 kHz.
Explanation:
Given that,
A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm
The speed of sound is, v = 336 m/s
We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :
So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.
