Answer:
Explanation:
For third normal mode of vibration
l = , λ is wavelength , l is length of string .
.4 =
λ = .267 m
velocity =
T is tension and m is mass unit length
m = .5 x 10⁻³ / 40 x 10⁻²
= .00125 kg / m
Putting the values
velocity =
= 253 m /s
frequency
= velocity / λ
= 253 / .267
= 947.5 Hz .
The answer is shape, waves keep the same shape when bouncing off something
Since no external torques are acting on the skater, his angular momentum must be conserved in the two different situations (arms and legs in - arms out and one leg out). The angular momentum is given by
where I is the moment of inertia and
the angular speed.
Labeling with "1" the situation where the skater has arms and legs in, and with "2" the situation where the skater has arms out and one leg out, the conservation of the angular momentum becomes
or
We know the moment of inertia and the angular speed of the skater when he has arms and legs in:
and also the moment of inertia of the skater when he has arms and one leg out:
so we can find his angular speed when he opens the arms and one leg:
Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C