Answer:
Amplitude = 0.058m
Frequency = 6.25Hz
Explanation:
Given
Amplitude (A) = 8.26 x 10-2 m
Frequency (f) = 4.42Hz
Conversation of energy before split
½mv² = ½KA²
Make A the subject of formula
A =
Conversation of energy after split
½(m/2)V'² = ½(m/2)V² = ½KA'²
½(m/2)V² = ½KA'²
Make A the subject of formula
First divide both sides by ½
(m/2)V² = KA'²
Divide both sides by K
V² = A'²
= A'
Substitute
for A in the above equation
A' = A/√2
A' = 8.26 x 10^-2/√2
A' = 0.05840702012600882
Amplitude after split = 0.058 (Approximated)
Frequency (f') = f√2
f' = 4.42√2
f' = 6.25082394568908011
Frequency after split = 6.25Hz (approximated)
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
As they evolved and adapted, those of the fittest survived to reproduce. Those that did not adapt, died.
neutrons
Explanation:
Isotopes of an element differ in their number of neutrons.
Isotopy is the existence of two or more atoms of the same element having the same atomic number but different mass numbers due to the differences in the number of neutrons in their nucleus.
- Isotopes of an element have the same electronic configuration.
- They have the same chemical properties but differ in their masses.
- The mass of an atom is function of the protons and neutrons.
Learn more:
Isotope brainly.com/question/2593342
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Answer:
Part(a): The frequency is
.
Part(b): The speed of the wave is
.
Explanation:
Given:
The distance between the crests of the wave,
.
The time required for the wave to laps against the pier, 
The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is
.
Also, the time required for the wave for each laps is the time period of oscillation and it is given by
.
Part(a):
The relation between the frequency and time period is given by

Substituting the value of
in equation (1), we have

Part(b):
The relation between the velocity of a wave to its frequency is given by

Substituting the value of
and
in equation (2), we have
