Answer:
1.) Frequency F = 890.9 Hz
2.) Wavelength (λ) = 0.893 m
Explanation:
1.) Given that the wavelength = 0.385m
The speed of sound = 343 m / s
To predict the frequency, let us use the formula V = F λ
Where (λ) = wavelength = 0.385m
343 = F × 0.385
F = 343/0.385
F = 890.9 Hz
2.) Given that the frequency = 384Hz
Using the formula again
V = F λ
λ = V/F
Wavelength (λ) = 343/384
Wavelength (λ) = 0.893 m
The two questions can be solved with the use of formula
Answer: the answer would be four thousand
Explanation: hope this helps
Answer:
a) 567J
b) 283.5J
c)850.5J
Explanation:
The expression for the translational kinetic energy is,
Substitute,
14kg for m
9m/s for v
The translational kinetic energy of the center of mass is 567J
(B)
The expression for the rotational kinetic energy is,
The expression for the moment of inertia of the cylinder is,
The expression for angular velocity is,
substitute
1/2mr² for I
and vr for w
in equation for rotational kinetic energy as follows:
The rotational kinetic energy of the center of mass is 283.5J
(c)
The expression for the total energy is,
substitute 567J for E(r) and 283.5J for E(R)
The total energy of the cylinder is 850.5J
Answer:
P.E. = -0.449 J
Explanation:
Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.
Now Potential energy is defined by this formula,
P.E. = k q₁ q₂/ r
where P.E. is the potential energy.
k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)
q₁ = charge of one particle = +1.0μC
q₂ = charge of another particle = -5.0μC
r = distance = 0.1 m
Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m
P.E. = -0.449 J
Ans: Dilute the solution
Explanation:
To decrease the over-saturation, dilute the solution. Dilution<span> is the process of decreasing the solute's concentration in the </span>solution. It is<span> usually done by mixing with more solvent. In other words, to </span>dilute<span> a </span>solution<span> means to add more solvent without the addition of more solute.</span>
