Explanation:
Given that,
2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.
The tension in one of them is then decreased sightly, then will decrese.
Beat frequency,
So, the new frequency of the string is 217 Hz. Hence, this is the required solution.
Answer:
The position on the x axis is 0.32 m.
Explanation:
Given that,
Point charge = 27 nC
Charge = 6 nC
Distance = 1
We need to calculate the distance
Using formula of electric field
Put the value into the formula
Hence, The position on the x axis is 0.32 m.
Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;
Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
Therefore, the extension of the wire is 0.362 mm.