Answer:
Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.
This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.
Answer:
Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...
Explanation:
Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.
For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc
If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc
Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...
Y₀ = initial position of the balloon at the top of the building = 44 m
Y = final position of the balloon at halfway down the building = 44/2 = 22 m
a = acceleration of the balloon = - 9.8 m/s²
v₀ = initial velocity of the balloon = 0 m/s
v = final velocity of the balloon = ?
using the kinematics equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
v² = 0² + 2 (- 9.8) (22 - 44)
v = 20.78 m/s
Answer:
The stirring allows fresh solvent molecules to continually be in contact with the solute. If it is not stirred, then the water right at the surface of the solute becomes saturated with dissolved sugar molecules, meaning that it is more difficult for the additional solute to dissolve.
Explanation: Help this helps :)))
Answer:
The potential energy (P.E) at the top is 392 J
The kinetic energy (K.E) at the top is 0 J
The potential energy (P.E) at the halfway point is 196 J.
The kinetic energy (K.E) at the halfway point is 196 J.
Explanation:
Given;
mass of the rock, m = 2 kg
height of the cliff, h = 20 m
speed of the rock at the halfway point, v = 14 m/s
The potential energy (P.E) and kinetic energy (K.E) when its at the top;
P.E = mgh
P.E = (2)(9.8)(20)
P.E= 392 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the top of the cliff = 0
K.E = ¹/₂(2)(0)²
K.E = 0
The potential energy (P.E) and kinetic energy (K.E) at the halfway point;
P.E = mg(¹/₂h)
P.E = (2)(9.8)(¹/₂ x 20)
P.E = 196 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the halfway point = 14 m/s
K.E = ¹/₂(2)(14)²
K.E = 196 J.
