<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source.

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>

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3 years ago

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A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface

Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so: