Answer:
OR ![H_2 =14.5 \ m)](https://tex.z-dn.net/?f=H_2%20%20%3D14.5%20%5C%20%20m%29)
Explanation:
From the question we are told that
The constant speed of the balloon is ![v = 6.60 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%206.60%20%5C%20m%2Fs)
The height of the balloon is ![h = 11.0 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%2011.0%20%5C%20%20m)
The initial speed of the pellet is ![u = 30 \ m/s](https://tex.z-dn.net/?f=u%20%20%3D%20%2030%20%5C%20m%2Fs)
Generally the height of the balloon at the point it is the same altitude with the pellet is mathematically represented as
![H = h + v * (t)](https://tex.z-dn.net/?f=H%20%20%3D%20%20h%20%20%2B%20%20v%20%2A%20%28t%29)
Note: vt is the distance covered by the balloon before the pellet got to it
Generally the height of the pellet when it is the same height with the balloon is mathematically represented using kinematics equation
![s = ut + \frac{1}{2} gt^2](https://tex.z-dn.net/?f=s%20%20%3D%20%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
So
H = s
=> ![ut + \frac{1}{2} gt^2 = h + v * (t)](https://tex.z-dn.net/?f=%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%20%3D%20%20h%20%20%2B%20%20v%20%2A%20%28t%29)
=> ![30t + \frac{1}{2} *( -9.8)t^2 = 11 + 6.60t](https://tex.z-dn.net/?f=%2030t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%28%20-9.8%29t%5E2%20%3D%20%2011%20%20%2B%20%206.60t)
=> ![4.9t^2 -23.4t + 11= 0](https://tex.z-dn.net/?f=%20%204.9t%5E2%20%20-23.4t%20%2B%2011%3D%20%200)
using the quadratic formula to solve the above equation
From the quadratic formula calculation
![t_1 = 4.25 \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%204.25%20%5C%20%20s)
OR
![t_1 = 0.529 \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%200.529%20%20%5C%20%20s)
So the height of this two place above the ground is mathematically evaluated as
![H_1 = h + v * (4.25)](https://tex.z-dn.net/?f=H_1%20%20%3D%20%20h%20%20%2B%20%20v%20%2A%20%284.25%29)
![H_1 = 11 + 6.60 * (4.25)](https://tex.z-dn.net/?f=H_1%20%20%3D%2011%20%20%2B%20%206.60%20%2A%20%284.25%29)
![H_1 =39.05 \ m](https://tex.z-dn.net/?f=H_1%20%20%3D39.05%20%5C%20m)
OR
![H_2 = h + v * ( 0.529)](https://tex.z-dn.net/?f=H_2%20%20%3D%20%20h%20%20%2B%20%20v%20%2A%20%28%200.529%29)
![H_2 = 11 + 6.60 * (0.529)](https://tex.z-dn.net/?f=H_2%20%20%3D%2011%20%20%2B%20%206.60%20%2A%20%280.529%29)
![H_2 =14.5 \ m)](https://tex.z-dn.net/?f=H_2%20%20%3D14.5%20%5C%20%20m%29)
Answer:
Yes, the frequency of light emitted is a property of the difference between the levels of energy of its electrons.
Explanation:
Neon atom is a noble gas which glows when its electrons de-excites after absorbing energy.
Niels Bohr postulated that the energy level in all atoms are quantized, thus electrons do not exist in-between two levels. When electrons in the Neon atom are excited, this increase in energy causes them to jump to a higher energy levels. On de-excitation, the electrons drops to their initial level releasing the absorbed energy in the form of a photon.
The photon emitted has a frequency that is directly proportional to the energy change in the electron.
(AB) & (CD)
Most machines are fueled by gasoline and electricity
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = (
-
) / (
+
)
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
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