In the context of variables and measurement scales, the magnitude of the differences between the numbers on the scale is meaningful interval scale.
<h3>What is an interval scale?</h3>
It should be noted that an interval scale means a measurement where the difference between the variables is meaningful and equal.
In this case, in the context of variables and measurement scales, the magnitude of the differences between the numbers on the scale is meaningful interval scale.
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Answer:
The distance can the water be projected is 4.51 m
Explanation:
The speed of the water in the hose is equal to:
v1 = R/A1
If we solve the continuity for v2:
v2 = R/A2 (eq. 1)
The equation for the vertical position is:
yf = yi + vy*t - (1/2)gt²
yi = 0
vy = 0
Clearing t:
(eq. 2)
The equation for position is:
xf = xi + vxt = 0 + v2t = v2t (eq. 3)
Replacing equation 1 and 2 in equation 3:

Answer:
B internal (thermal) energy
Carson's Rule says:
FM occupied bandwidth =
(2) · (Peak deviation + Highest modulating frequency)
so
FM bw = (2) · (60 kHz + 15 kHz)
FM bw = (2) · (75 kHz)
FM bw = 150 kHz
(I used to eat this stuff for lunch, but it's been almost 40 years. Thanks for taking me back. Those were the good old days.)
By power transmitted by string, the frequency must be 163.31 Hz.
We need to know about power transmitted to solve this problem. The power transmitted by a wave on string can be determined by this equation
P = 1/2 . μ . ω² . A² . v
where P is the power, μ is mass per unit length of string, ω is angular speed, A is amplitude and v is wave propagation speed.
the wave propagation can be determined as
v = √(F.l/m)
where F is the string tension, l is length and m is the mass.
From the question above, we know that:
l = 2.7 m
m = 260 g = 0.26 kg
F = 36 N
A = 7.7 mm = 0.0077 m
P = 58 W
Find the mass per unit length
μ = m / l
μ = 0.26 / 2.7
μ = 0.096 kg / m
Find the wave propagation speed
v = √(F.l/m)
v = √(36. 2.7 /0.26)
v = √(373.85)
v = 19.34 m/s
Find the angular speed
P = 1/2 . μ . ω² . A² . v
58 = 1/2 . 0.096 . ω² . 0.0077² . 19.34
ω² = 1053777.29
ω = √1053777.29
ω = 1026.54 rad/s
Find the frequency
ω = 2πf
1026.54 = 2 . 3.14 . f
f = 163.31 Hz
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