Two compact sources of sound oscillate in phase with a frequency of 100 Hz. At a point 5.00m from one source and 5.85 m from the
1 answer:
You might be interested in
Answer:
Q = 1.52 X 10⁻⁷ C.
Explanation:
String is inclined at 10 degree with the vertical so the vertical component of T will balance the weight and the horizontal component will balance the force of repulsion F.
T Cos 10 = mg
T Sin 10 = F
Tan 10 = F / mg
F = mg Tan 10
= 10⁻³ x 9.8 x .176
= 1.728 x 10⁻³ .
Force of repulsion F = K Q² / R²
Distance between two charges = 2 x 1 x sin 10
= .347 m
1.728 X 10⁻³ = [/tex]
Q² = 2.31 X 10⁻¹⁴
Q = 1.52 X 10⁻⁷ C.
The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.
Given:
mass of block 1, m₁ = 40 gm = 40×10⁻³ kg
mass of block 2, m₂ = 20 gm = 20×10⁻³ kg
Applied force, F = 0.03 N
Calculation:
Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:
F = ma
where F is the applied force
m is the total mass of the system
a is the acceleration of block 2 (as it is pulled by horizontal force)
From the above equation we get:
0.03 N = (m₁+m₂) a
⇒ a = (0.03 N) / (m₁+m₂)
⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)
⇒ a = 0.5 m/s²
Now, from the free-body diagram of block 2 as shown in figure 3, we get:
Balancing the forces along the horizontal:
∑Fₓ = 0
∴ T = m₂ a
where T is tension in the string
a is the acceleration of block 2
Applying values in the above equation we get:
T = (20×10⁻³ kg) × (0.5 m/s²)
= 0.01 N
Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.
Learn more about tension here:
<u>brainly.com/question/13397436</u>
#SPJ4
