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mrs_skeptik [129]
3 years ago
11

What does the person in back of the ambulance experience? A) a lower frequency of the siren B) a lower amplitude of the siren C)

a higher frequency of the siren Eliminate D) a higher amplitude of the siren
Physics
2 answers:
hoa [83]3 years ago
5 0

The answer to your question is "A. a lower frequency of the siren.

Because the person in back of the ambulance will hear a lower frequency of the siren. This is because the waves are stretched out. A longer wavelength results in a lower frequency.

denpristay [2]3 years ago
4 0

Answer: The correct answer is A.

Explanation:

When there is a relative motion between the source and the observer then there will be change in the frequency or wavelength of the wave. This phenomenon is called Doppler effect.

For example, when the source and the observer are moving away from other then the frequency decreases in this case.

In the given problem, the person is in the back of the ambulance. In this case, the person in the back of the ambulance will listen a lower frequency of the wave. There will be longer wavelength.

Therefore, the correct answer is A.

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A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
Elanso [62]
F=55 \ N \\ d=10 \ m \\ \boxed{W-?} \\ \bold{Solving:} \\ \boxed{W=F \cdot d} \\ W=55\ N \cdot 10 \ m \\ \Rightarrow \boxed{W=550 \ J}
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5. A car accelerates from rest down a highway at 6 m/s2 for 5 seconds. What distance did the car travel during acceleration?
Mumz [18]

Answer:

75 m

Explanation:

The distance x for a uniform acceleration is given by the equation:

x = \frac{1}{2}at^{2}+v_ot+x_o

The given values are:

a = 6 m/s², t = 5 s, v₀ = 0, x₀ = 0

5 0
3 years ago
The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I
erma4kov [3.2K]

Answer:

Part a)

I_{end} = \frac{mL^2}{3}

Part b)

I_{edge} = \frac{2ma^2}{3}

Explanation:

As we know that by parallel axis theorem we will have

I_p = I_{cm} + Md^2

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

I = \frac{mL^2}{12}

now if we need to find the inertia from its end then we will have

I_{end} = I_{cm} + Md^2

I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2

I_{end} = \frac{mL^2}{3}

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

I = \frac{ma^2}{6}

now if we need to find the inertia about an axis passing through its edge

I_{edge} = I_{cm} + Md^2

I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2

I_{edge} = \frac{2ma^2}{3}

7 0
3 years ago
A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
algol13

Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

Using formula of linear density

\mu_{s}=\rho_{s}A

\mu_{s}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{s}=5.00\times10^{-3}\ kg/m

We need to calculate the linear density of copper wire

Using formula of linear density

\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{c}=5.67\times10^{-3}\ kg/m

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}

v_{s}=170.3\ m/s

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

v_{c}=159.9\ m/s

We need to calculate the time taken by the wave to travel  along the combination of two wires

t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

4 0
3 years ago
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