Answer:
The speed of sound in the argon is 340 m/s.
Explanation:
Given;
fundamental frequency, f₀ = 200 Hz
length of the pipe, L = 42.5 cm = 0.425 m
A pipe that is open at one end and closed at another end is known as a closed pipe.
The wavelength for the first harmonic is calculated as;
L = Node -------> Antinode
L = λ/4
λ = 4L
The speed of the sound is calculated as;
v = fλ
where;
v is the speed of the sound
v = f x 4L
v = 200 x (4 x 0.425)
v = 340 m/s
Therefore, the speed of sound in the argon is 340 m/s.
Answer:
The spring constant is 60,000 N
The total work done on it during the compression is 3 J
Explanation:
Given;
weight of the girl, W = 600 N
compression of the spring, x = 1 cm = 0.01 m
To determine the spring constant, we apply hook's law;
F = kx
where;
F is applied force or weight on the spring
k is the spring constant
x is the compression of the spring
k = F / x
k = 600 / 0.01
k = 60,000 N
The total work done on the spring = elastic potential energy of the spring, U;
U = ¹/₂kx²
U = ¹/₂(60000)(0.01)²
U = 3 J
Thus, the total work done on it during the compression is 3 J
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
Answer:
B. 2
Explanation:
The reaction expression is given as:
_S + 3O₂ → 2SO₃
Now let us balanced the expression;
On the product side we have 2 moles of S
On the reactant side we should have 2moles of S
So, we put the coefficient 2 to balance the expression;
We have 6 moles oxygen on both sides
Answer:
I think the acceleration is 12m/s
