Answer:
f= 440.4 Hz or f= 439.6 Hz
Explanation:
In this exercise we have two waves with slightly different frequencies, the A4 chord with f = 440 Hz and the beating with the central chord fbeats = 0.40 Hz, let's use the relation
f_{beat} = | f - f₀ |
f = fo + - f _{beat}
f = 440 + 0.4 = 440.4 Hz
f = 440-0.4 = 439.6 Hz
I believe it’s a, hope it helps on me
Answer:
42.5 m/s
Explanation:
Given:
x₀ = 0 m
x = 62 m
y₀ = 80 m
y = 0 m
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
Find: v
First, find the time it takes to land.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (80 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 4.04 s
Find the horizontal component vₓ:
x = x₀ + vₓ t − ½ aₓ t²
(62 m) = (0 m) + vₓ (4.04 s) − ½ (0 m/s²) (4.04 s)²
vₓ = 15.3 m/s
Find the vertical component vᵧ:
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8 m/s²) (4.04 s) + (0 m/s)
vᵧ = -39.6 m/s
Find the speed using Pythagorean theorem:
v = √(vₓ² + vᵧ²)
v = √((15.3 m/s)² + (-39.6)²)
v = 42.5 m/s
Answer:
dt/dx = -0.373702
dt/dy = -1.121107
Explanation:
Given data
T(x, y) = 54/(7 + x² + y²)
to find out
rate of change of temperature with respect to distance
solution
we know function
T(x, y) = 54 /( 7 + x² + y²)
so derivative it x and y direction i.e
dt/dx = -54× 2x / (7 +x² + y²)² .........................1
dt/dy = -54× 2y / (7 + x² + y²)² .........................2
now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2
dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²
dt/dx = -0.373702
and
dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²
dt/dy = -1.121107
