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:
a = -7.29 m / s²
Explanation:
For this exercise we must use Newton's second law,
F -W = m a
Force is electrical force
F = k q₁ q₂ / r²
k q₁ q₂ / r² -mg = m a
indicate that the charge of the two spheres is equal
q₁ = q₂ = q
a = (k q² / r² - m g) / m
a = k q² / m r² - g
Let's reduce the magnitudes to the SI system
m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg
q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C
r = 10.0 cm (1m / 100cm) = 0.1000 m
let's calculate
a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8
a = -7.29 m / s²
The negative sign indicates that the direction of this acceleration is downward
Answer:

So a=3.844 and b=5
Explanation:
Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

with a=3.844, and "5" as the exponent of ten (so b=5)
Answer:
there will be collision
Explanation:
= speed of sue = 34 m/s
= speed of van = 5.20 m/s
= speed of sue relative to van =
= 34 - 5.20 = 28.8 m/s
= stopping distance after brakes are applied
= distance between sue and van = 160 m
= final speed of sue = 0 m/s
= acceleration = - 1.80 m/s²
Using the kinematics equation


m
Since
hence there will be collision