Answer:
A radio telescope helped the astronomers discover the CMB.
Explanation:
- Penzias and Wilson while experimenting with a radio telescope in 1964, accidentally discovered the radiation that exists universally also known as the CMB.
- This was used to support the "Big Bang Theory" and not the "Steady State Theory"
- CMB is the faint cosmic radiation that fills up the universe. It provides important data for understanding early universe.
- This data tells us about the composition of the universe and its age which raises new questions about the universe.
"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"
In series circuit, Req = R₁ + R₂ + R₃ + ···
In parallel circuit, 
<h3>Q7.</h3>
total resistance in the upper branch = R₂ + R₃ = R₂ + 2


R₂ + 2 = 12
R₂ = 10Ω
<h3>Q8.</h3>


Req = 1.7Ω
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;

E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C