Answer:
The beat frequency is 30 Hz
Explanation:
Given;
velocity of the two sound waves, v = 343 m/s
wavelength of the first wave, λ₁ = 5.72 m
wavelength of the second wave, λ₂ = 11.44 m
The frequency of the first wave is calculated as follows;
F₁ = v/λ₁
F₁ = 343 / 5.72
F₁ = 59.97 HZ
The frequency of the second wave is calculated as follows;
F₂ = v/λ₂
F₂ = 343 / 11.44
F₂ = 29.98 Hz
The beat frequency is calculated as;
Fb = F₁ - F₂
Fb = 59.97 HZ - 29.98 Hz
Fb = 30 Hz
The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
Learn more about Conservation of Linear Momentum here:
brainly.com/question/7538238
Answer:

Explanation:
It is given that,
Length of the wire, L = 0.6 m
Current flowing inside the wire, I = 2 A
Uniform magnetic field, B = 0.3 T
Force experienced by the wire in the magnetic field, F = 0.18 N
To find,
The angle made by the wire with the magnetic field.
Solve,
We know that the magnetic force acting on the wire inside the magnetic field is given by :




Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Answer:
<u><em>Electric Potential Energy:</em></u>
The energy that is needed to move a charge against an electric firld is called Electric Potential Energy
<u><em>Electric Potential Difference:</em></u>
The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.
<u><em>Relation:</em></u>
Relation between Electric potential and electrical potential energy is given by

Here PE represents Electric potential energy
and
is Electric potential difference
it means electric potential difference is the difference in electric potential energy divided by the charge.