The answer is in the attachment
<span>...........................................</span>
<span>...........................................</span>
A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 30.0 m/s. If the s
1 answer:
Answer:
548.5 Hz
Explanation:
We are given that
Frequency of source=500 Hz
Speed of train=30m/s
Speed of sound in air=339m/s
We have to find the frequency of the bell to an observer riding the train.
The apparent frequency for the listener
Where Frequency of source
=Speed of sound
=Speed of source
Using the formula
The apparent frequency of the bell to an observer riding the train=
Hence, the apparent frequency of the bell to an observer riding the train=548.5 Hz
You might be interested in
Answer:
a) E =0, b) E = 1,129 10¹⁰ N / C , c) E = 3.33 10¹⁰ N / C
Explanation:
To solve this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.
E A = /ε₀
The area of a sphere is
A = 4π r²
E = q_{int} / 4πε₀ r²
k = 1 / 4πε₀
E = k q_{int} / r²
To find the charge inside the surface we can use the concept of density
ρ = q_{int} / V ’
q_{int} = ρ V ’
V ’= 4/3 π r’³
Where V ’is the volume of the sphere inside the Gaussian surface
Let's apply this expression to our problem
a) The electric field in center r = 0
Since there is no charge inside, the field must be zero
E = 0
b) for the radius of r = 6.0 cm
In this case the charge inside corresponds to the inner sphere
q_{int} = 5.0 4/3 π 0.06³
q_{int} = 4.52 10⁻³ C
E = 8.99 10⁹ 4.52 10⁻³ / 0.06²
E = 1,129 10¹⁰ N / C
c) The electric field for r = 12 cm = 0.12 m
In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.
The charge of the inner sphere is q₁ = - 4.52 10⁻³ C
The charge for the outermost sphere is
q₂ = ρ 4/3 π r₂³
q₂ = 8.0 4/3 π 0.12³
q₂ = 5.79 10⁻² C
The net charge is
q_{int} = q₁ + q₂
q_{int} = -4.52 10⁻³ + 5.79 10⁻²
q_{int} = 0.05338 C
The electric field is
E = 8.99 10⁹ 0.05338 / 0.12²
E = 3.33 10¹⁰ N / C
Answer:
"0.049 W" is the correct answer.
Explanation:
According to the given question,
As we know,
⇒
Now,
⇒
or,
Answer:
Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-
Substituting the given values in above equation, we get -
Now we will determine the radius of curvature for the curve shown in the attached image
Differentiating on both the sides, we get -
meter
Acceleration on curved path
Final acceleration
