Answer:
The frequency of the second harmonic (
) is 11.97 Hz.
Explanation:
Given;
mass of the string, m = 25 g = 0.025kg
tension on the string, T = 43 N
length of the string, L = 12 m
The speed of wave on the string is given as;
where;
μ is mass per unit length = 0.025 / 12 = 0.002083 kg/m
The wavelength of the first harmonic wave is given as;
The frequency of the first harmonic is given as;
The wavelength of the second harmonic wave is given as;
The frequency of the second harmonic is given as;
Therefore, the frequency of the second harmonic () is 11.97 Hz.
The correct answer to the question is- C).Electrons are ripped off one material and held tightly by the other material.
EXPLANATION:
Before going to answer this question, first we have to understand the charging by friction.
There are three modes of charging a body, which are known as friction,conduction, and induction.
Friction: Friction is the type of charging a body due to the actual transfer of electrons when two substances of different electron affinities are rubbed with each other. During rubbing, the electrons from less electron affinity substance will be transferred to high affinity substance.
The substance which will accept electrons gets negatively charged, and the other substance will be positively charged.
There will be no transfer of protons from one substance to another as protons are bound inside the nuclei of atoms.
Hence, the correct answer to the question is that electrons are ripped off one material and held tightly by the other material.
The answer is “B” hope I helped
Answer:
For these reasons at 98 mph the path is straighter
Explanation:
To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels
X = Vox t
Y = Yo + Voy t - ½ g t²
They tell us that the only parameter that changes is the speed, so the distance to the plate is known
t = Vox / x
We replace
Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2
Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²
Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)
The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.
Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter
