Answer:
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Explanation:
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Dehydration? I think that’s it.
The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as

Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as

Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well


Replacing
for n and 202Hz for 



The frequency of the tightened is 205Hz
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
The acceleration exerted by the object of mass 10 kg is 
Answer: Option A
<u>Explanation:</u>
According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

So,

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

So, the acceleration exerted by the object of mass 10 kg is 