Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
This question is a bit ambiguous because all parts of a scientific argument must be supported by valid data. However, among the choices, the closest synonym to "valid data" would be evidence. Evidence is the body of facts or information that support a given idea.
Answer:
procedure in which radio waves and a powerful magnet linked to a computer
Explanation:
Listen to pronunciation. (mag-NEH-tik REH-zuh-nunts IH-muh-jing) A procedure in which radio waves and a powerful magnet linked to a computer are used to create detailed pictures of areas inside the body. These pictures can show the difference between normal and diseased tissue.
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.
The Coulomb force between the two balls is:
where
is the Coulomb's constant,
is the intensity of the two charges, and
is the distance between them.
Substituting these numbers into the equation, we get
The force is repulsive, because the charges have same sign and so they repel each other.
When the mass of the cart changes, the time to travel at 4.6 m/s is 28.11 s.
<h3>
Acceleration of the mule</h3>
The acceleration of mule is calculated as follows;
a = v/t
a = 5/10
a = 0.5 m/s²
<h3>For constant applied force</h3>
F1 = F2
m₁v₁/t₁ = m₂v₂/t₂
(180 x 5) / 10 = (550 x 4.6)/t
90 = 2530/t
t = 2530/90
t = 28.11 s
Thus, when the mass of the cart changes, the time to travel at 4.6 m/s is 28.11 s.
Learn more about acceleration here: brainly.com/question/14344386
